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Copyright © Big Ideas Learning, LLC Algebra 2 289All rights reserved. Worked-Out Solutions
Chapter 6
Chapter 6 Maintaining Mathematical Profi ciency (p. 293)
1. 3 ⋅ 24 = 3 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2
= 3 ⋅ 16
= 48
2. (−2)5 = (−2)(−2)(−2)(−2)(−2)
= −32
3. − ( 5 — 6 )
2
= − ( 5 — 6 ) ( 5 —
6 )
= − 25 —
36
4. ( 3 — 4 )
3
= ( 3 — 4 ) ( 3 —
4 ) ( 3 —
4 )
= 27 —
64
5. The domain of the function is −5 ≤ x ≤ 5 and the range is
0 ≤ y ≤ 5.
6. The domain of the function is {−2, −1, 0, 1, 2} and the
range is {−2, −1, 0, 1, 2}.
7. The domain of the function is all real numbers and the range
is y ≤ 0.
8. The expression −4n = −1 ⋅ (4n) is negative for all values
of n and the expression (−4)n = (−1)n ⋅ (4n) is negative
when n is odd. The expression −4n is never positive and the
expression (−4)n is positive when n is even.
Chapter 6 Mathematical Practices (p. 294)
1. Begin by determining whether the ratios of consecutive
y-values are equal.
2 —
1 = 2,
4 —
2 = 2,
8 —
4 = 2,
16 —
8 = 2
The ratios of consecutive y-values are equal, so the data can
be modeled by
y = 2x.
Using the model, when x = 10, y = 210 = 1024.
2. Because the fi rst point is (0, 0), fi rst determine whether the
differences of consecutive y-values are equal.
4 − 0 = 4, 8 − 4 = 4, 12 − 8 = 4, 16 − 12 = 4 The differences are equal, so the data can be modeled by
y = 4x.
Using this model, when x = 10, y = 4(10) = 40.
3. Begin by determining that the ratios of consecutive y-values
are equal.
4 —
1 = 4,
7 —
4 =
7 —
4 ,
10 —
7 =
10 —
7 ,
13 —
10 =
13 —
10
The ratios of consecutive y-values are not equal, so the
data cannot be modeled by an exponential function. Next,
determine whether the differences of consecutive y-values
are equal.
4 − 1 = 3, 7 − 4 = 3, 10 − 7 = 3, 13 − 10 = 3 The differences are equal, so the data can be modeled by
y = 3x + 1.
Using the model, when x = 10, y = 3(10) + 1 = 31.
4. Begin by determining whether the ratios of consecutive
y-values are equal.
3 —
1 = 3,
9 —
3 = 3,
27 —
9 = 3,
81 —
27 = 3
The ratios of consecutive y-values are equal, so the data can
be modeled by
y = 3x.
Using the model, when x = 10, y = 310 = 59,049.
6.1 Explorations (p. 295)
1. a. x −2 −1 0 1 2
f (x) = 2x 1 —
4
1 —
2 0 2 4
The function matches graph A.
b. x −2 −1 0 1 2
f (x) = 3x 1 —
9
1 —
3 1 3 9
The function matches graph F.
c. x −2 −1 0 1 2
f (x) = 4x 1 —
16
1 —
4 1 4 16
The function matches graph C.
d. x −2 −1 0 1 2
f (x) = ( 1 — 2 ) x
4 2 1 1 —
2
1 —
4
The function matches graph D.
e. x −2 −1 0 1 2
f (x) = ( 1 — 3 ) x
9 3 1 1 —
3
1 —
9
The function matches graph B.
f. x −2 −1 0 1 2
f (x) = ( 1 — 4 ) x
16 4 1 1 —
4
1 —
16
The function matches graph E.
290 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
2. The domain of f (x) = bx is all real numbers because the
graphs have no holes, asymptotes, or stopping points. The
range is y > 0 because the graphs approach but do not touch
the x-axis. The y-intercept is 1 because f (0) = b0 = 1.
3. The domain is all real numbers, the range is y > 0, and the
x-axis is an asymptote.
4. It is not possible for the graph of f (x) = bx to have an
x-intercept because when b > 0, bx > 0.
6.1 Monitoring Progress (pp. 297–299)
1. Step 1 Identify the value of the base. The base, 4, is greater
than 1, so the function represents exponential growth.
Step 2 Make a table of values.
x −1 − 1 — 2 0
1 —
2 1
3 —
2
y 1 —
4
1 —
2 1 2 4 8
Step 3 Plot the points from the table.
Step 4 Draw, from left to right, a smooth curve that begins
just above the x-axis, passes through the plotted
points, and moves up to the right.
y
42−2−4
2
4
6
8
x
2. Step 1 Identify the value of the base. The base, 2 —
3 , is greater
than 0 and less than 1, so the function represents
exponential decay.
Step 2 Make a table of values.
x −3 −2 −1 0 1 2
y 27
— 8
9 —
4
3 —
2 1
2 —
3
4 —
9
Step 3 Plot the points from the table.
Step 4 Draw, from right to left, a smooth curve that begins
just above the x-axis, passes through the plotted
points, and moves up to the left.
y
42−2−4
2
4
6
8
x
3. Step 1 Identify the value of the base. The base, 0.25,
is greater than 0 and less than 1, so the function
represents exponential decay.
Step 2 Make a table of values.
x −1.5 −1 −0.5 0 0.5 1
y 8 4 2 1 0.5 0.25
Step 3 Plot the points from the table.
Step 4 Draw, from right to left, a smooth curve that begins
just above the x-axis, passes through the plotted
points, and moves up to the left.
y
42−2−4
4
6
8
x
4. Step 1 Identify the value of the base. The base,
1.5, is greater than 1, so the function represents
exponential growth.
Step 2 Make a table of values.
x −2 −1 0 1 2 3
y 4 —
9
2 —
3 1
3 —
2
9 —
4
27 —
8
Step 3 Plot the points from the table.
Copyright © Big Ideas Learning, LLC Algebra 2 291All rights reserved. Worked-Out Solutions
Chapter 6
Step 4 Draw, from left to right, a smooth curve that begins
just above the x-axis, passes through the plotted
points, and moves up to the right.
y
42−2−4
2
4
6
8
x
5. The decay factor is 0.9 = 1 − 0.1, so the annual percent
decrease is 0.1 or 10%. Use a graphing calculator to
determine that y = 8000 when t ≈ 10.8. After about
10.8 years, the value of the car will be about $8000.
6. The equation is
y = 6.09(1 + 0.015)t
= 6.09(1.015)t.
Use a graphing calculator to determine that y ≈ 7 when t ≈ 9.
So, the world population was about 7 billion in 2009.
7. y = a(0.5)t/13
= a[(0.5)1/13]t
≈ a(0.9481)t
= a(1 − 0.0519)t
The hourly decay rate is about 0.0519, or 5.19%.
8. A = P ( 1 + r — n )
nt
= 9000 ( 1 + 0.0146 —
365 ) 365
⋅ 3
≈ 9402.95
The balance at the end of 3 years is $9402.95.
6.1 Exercises (pp. 300–302)
Vocabulary and Core Concept Check
1. The initial amount is 2.4, the growth factor is 1.5, and
the percent increase is 0.5, or 50%.
2. The 80% decrease does not belong because it gives a
base of 0.2 rather than 0.8.
Monitoring Progress and Modeling with Mathematics
3. a. 2−2 = 1 — 4
b. 23 = 8
4. a. 4−2 = 1 — 16
b. 43 = 64
5. a. 8 ⋅ 3−2 = 8 ⋅ 1 —
9 =
8 —
9
b. 8 ⋅ 33 = 8 ⋅ 27 = 216
6. a. 6 ⋅ 2−2 = 6 ⋅ 1 — 4 = 3 —
2
b. 6 ⋅ 23 = 6 ⋅ 8 = 48
7. a. 5 + 3−2 = 5 + 1 — 9 = 46
— 9
b. 5 + 33 = 5 + 27 = 32
8. a. 2−2 − 2 = 1 — 4 − 2 = −
7 —
4
b. 23 − 2 = 8 − 2 = 6
9. Step 1 Identify the value of the base. The base, 6, is greater
than 1, so the function represents exponential growth.
Step 2 Make a table of values.
x −2 −1 0 1
y 1 —
36
1 —
6 1 6
Step 3 Plot the points from the table.
Step 4 Draw, from left to right, a smooth curve that begins
above the x-axis, passes through the plotted points,
and moves up to the right.
y
42−2−4
2
4
6
8
x
292 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
10. Step 1 Identify the value of the base. The base, 7, is greater
than 1, so the function represents exponential growth.
Step 2 Make a table of values.
x −2 −1 0 1
y 1 —
49
1 —
7 1 7
Step 3 Plot the points from the table.
Step 4 Draw, from left to right, a smooth curve that begins
above the x-axis, passes through the plotted points,
and moves up to the right.
y
42−2−4
2
4
6
8
x
11. Step 1 Identify the value of the base. The base, 1 —
6 , is greater
than 0 and less than 1, so the function represents
exponential decay.
Step 2 Make a table of values.
x −1 0 1 2
y 6 1 1 —
6
1 —
36
Step 3 Plot the points from the table.
Step 4 Draw, from right to left, a smooth curve that begins
just above the x-axis, passes through the plotted
points, and moves up to the left.
y
42−2−4
4
6
8
x
12. Step 1 Identify the value of the base. The base, 1 —
8 , is greater
than 0 and less than 1, so the function represents
exponential decay.
Step 2 Make a table of values.
x −1 0 1 2
y 8 1 1 —
8
1 —
64
Step 3 Plot the points from the table.
Step 4 Draw, from right to left, a smooth curve that begins
just above the x-axis, passes through the plotted
points, and moves up to the left.
y
42−2−4
6
8
x
13. Step 1 Identify the value of the base. The base, 4 —
3 , is greater
than 1, so the function represents exponential growth.
Step 2 Make a table of values.
x −2 −1 0 1 2 3
y 9 —
16
3 —
4 1
4 —
3
16 —
9
64 —
27
Step 3 Plot the points from the table.
Step 4 Draw, from left to right, a smooth curve that begins
just above the x-axis, passes through the plotted
points, and moves up to the right.
y
4 6 82−2
2
4
6
8
x
10
Copyright © Big Ideas Learning, LLC Algebra 2 293All rights reserved. Worked-Out Solutions
Chapter 6
14. Step 1 Identify the value of the base. The base, 2 —
5 , is greater
than 0 and less than 1, so the function represents
exponential decay.
Step 2 Make a table of values.
x −2 −1 0 1 2
y 25
— 4
5 —
2 1
2 —
5
4 —
25
Step 3 Plot the points from the table.
Step 4 Draw, from right to left, a smooth curve that begins
just above the x-axis, passes through the plotted
points, and moves up to the left.
y
42−2−4
4
6
8
x
15. Step 1 Identify the value of the base. The base, 1.2, is
greater than 1, so the function represents exponential
growth.
Step 2 Make a table of values.
x −2 −1 0 1 2 3
y 0.69 — 4 0.8
— 3 1 1.2 1.44 1.728
Step 3 Plot the points from the table.
Step 4 Draw, from left to right, a smooth curve that begins
just above the x-axis, passes through the plotted
points, and moves up to the right.
y
8 12 164−4
4
8
6
x
20
16
12
16. Step 1 Identify the value of the base. The base, 0.75,
is greater than 0 and less than 1, so the function
represents exponential decay.
Step 2 Make a table of values.
x −3 −2 −1 0 1 2
y 2. — 370 1.
— 7 1. — 3 1 0.75 0.5625
Step 3 Plot the points from the table.
Step 4 Draw, from right to left, a smooth curve that begins
just above the x-axis, passes through the plotted
points, and moves up to the left.
y
42−2−4−6−8
2
4
6
8
x
10
12
17. Step 1 Identify the value of the base. The base, 0.6, is
greater than 0 and less than 1, so the function
represents exponential decay.
Step 2 Make a table of values.
x −2 −1 0 1 2 3
y 2. — 7 1.
— 6 1 0.6 0.36 0.216
Step 3 Plot the points from the table.
Step 4 Draw, from right to left, a smooth curve that begins
just above the x-axis, passes through the plotted
points, and moves up to the left.
y
42−2−4
2
4
6
8
x
294 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
18. Step 1 Identify the value of the base. The base, 1.8, is
greater than 1, so the function represents exponential
growth.
Step 2 Make a table of values.
x −1 0 1 2 3
y 0. — 5 1 1.8 3.24 5.832
Step 3 Plot the points from the table.
Step 4 Draw, from left to right, a smooth curve that begins
just above the x-axis, passes through the plotted
points, and move up to the right.
y
42−2−4
2
4
6
8
x
19. The base b is 3 because the value of the function is 3 when
x = 1.
20. The base b is 5 because the value of the function is 5 when
x = 1.
21. a. The base, 0.75, is greater than 0 and less than 1, so the
model represents exponential decay.
b. Because t is given in years and the decay factor is
0.75 = 1 − 0.25, the annual percent decrease is 0.25,
or 25%.
c. Use the trace feature of a graphing calculator to determine
that y ≈ 50 when t = 4.8. After 4.8 years, the value of the
bike will be about $50.
22. a. The base, 1.03, is greater than 1, so the model represents
exponential growth.
b. Because t is given in years and the growth factor is
1.03 = 1 + 0.03, the annual percent increase is 0.03,
or 3%.
c. Use the trace feature of a graphing calculator to determine
that y ≈ 590 when t = 6. The population was about
590,000 six years after the beginning of the decade.
23. a. The initial amount is a = 233, and the percent increase
is r = 0.06. So, the exponential growth model is
y = a(1 + r)t
= 233(1 + 0.06)t
= 233(1.06)t.
Using this model, you can determine the number of cell
phone subscribers in 2008, when t = 2, to be
y = 233(1.06)2 ≈ 261.8 million.
b. Use the table feature of a graphing calculator to determine
that y ≈ 278 when t = 3. So, there were about 278 million
cell phone subscribers in 2009.
24. a. The initial amount is a = 325, and the percent decrease
is r = 0.29. So, the exponential decay model is
y = a(1 − r)t
= 325(1 − 0.29)t
= 325(0.71)t
b. Use the table feature of a graphing calculator to determine
that y ≈ 100 when t = 3.4. So, after about 3.4 hours there is
about 100 milligrams of ibuprofen in your blood stream.
25. y = a(3)t/14 Write original function.
= a[(3)1/14]t Power of a Power Property.
≈ a(1.0816)t Evaluate power.
= a(1 + 0.0816)t Rewrite in form y = a(1 + r)t.
26. y = a(0.1)t/3 Write original function.
= a[(0.1)1/3]t Power of a Power Property.
≈ a(0.4642)t Evaluate power.
= a(1 − 0.5358)t Rewrite in form y = a(1 − r)t.
27. y = a(0.5)t/5730
= a[(0.5)1/5730]t
≈ a(0.9999)t
= a(1 − 0.0001)t
The yearly decay rate is about 0.0001, or 0.01%.
28. y = a(1230.25)t/16
= a[(1230.25)1/16]t
≈ a(1.56)t
= a(1 + 0.56)t
The daily growth rate is about 0.56, or 56%.
29. y = a(2)t/3
= a[(2)1/3]t
≈ a(1.26)t
= a(1 + 0.26)t
The growth rate is about 0.26, or 26%.
Copyright © Big Ideas Learning, LLC Algebra 2 295All rights reserved. Worked-Out Solutions
Chapter 6
30. y = a(4)t/6
= a[(4)1/6]t
≈ a(1.26)t
= a(1 + 0.26)t
The growth rate is about 0.26, or 26%.
31. y = a(0.5)t/12
= a[(0.5)1/12]t
≈ a(0.94)t
= a(1 − 0.06)t
The decay rate is about 0.06, or 6%.
32. y = a(0.25)t/9
= a[(0.25)1/9]t
≈ a(0.86)t
= a(1 − 0.14)t
The decay rate is about 0.14, or 14%
33. y = a ( 2 — 3 )
t/10
= a [ ( 2 — 3 )
1/10
] t
≈ a(0.96)t
= a(1 − 0.04)t
The decay rate is about 0.04, or 4%
34. y = a ( 5 — 4 )
t/22
= a [ ( 5 — 4 )
1/22
] t
≈ a(1.01)t
= a(1 + 0.01)t
The growth rate is about 0.01, or 1%.
35. y = a(2)8t
= a[(2)8]t
= a(256)t
= a(1 + 255)t
The growth rate is 255, or 25,500%.
36. y = a ( 1 — 3 )
3t
= a [ ( 1 — 3 )
3
] t
≈ a(0.04)t
= a(1 − 0.96)t
The decay rate is about 0.96, or 96%.
37. With interest compounded quarterly (4 times per year), the
balance after 5 years is
A = P ( 1 + r — n ) nt
= 5000 ( 1 + 0.0225 —
4 ) 4 ⋅ 5
≈ 5593.60.
The balance at the end of 5 years is $5593.60.
38. Account Compounding Balance after 6 years
1 quarterly A = 2200 ( 1 + 0.03 —
4 ) 4 ⋅ 6
≈ 2632.11
2 monthly A = 2200 ( 1 + 0.03 —
12 ) 12
⋅ 6
≈ 2633.29
3 daily A = 2200 ( 1 + 0.03 —
365 ) 365
⋅ 6
≈ 2633.86
Account 1 earns about $2632.11 − $2200 = $432.11
in interest.
Account 2 earns about $2633.29 − $2200 = $433.29
in interest.
Account 3 earns about $2633.86 − $2200 = $433.86
in interest.
39. The percent decrease was used as the decay factor. The
decay factor is 1 − 0.02 = 0.98.
y = (initial amount)(decay factor)t
y = 500(0.98)t
40. The percent was not changed to its decimal form of 0.0125.
A = 250 ( 1 + 0.0125 —
4 ) 4 ⋅
3
A = $259.54
41. A = P ( 1 + r — n ) nt
= 3500 ( 1 + 0.0216 —
4 ) 4 ⋅ 6
= $3982.92
42. A = P ( 1 + r — n ) nt
= 3500 ( 1 + 0.0229 —
12 ) 12
⋅ 6
= $4014.98
43. A = P ( 1 + r — n )
nt
= 3500 ( 1 + 0.0183 —
365 ) 365
⋅ 6
= $3906.18
296 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
44. A = P ( 1 + r — n )
nt
= 3500 ( 1 + 0.0126 —
12 ) 12
⋅ 6
= $3774.71
45. The value of a represents the initial number of referrals.
So, the website received 2500 referrals at the start of the
10-year period. The value of b represents the growth factor.
The growth factor is 1.50 = 1 + 0.50, so the annual percent
increase is 0.50, or 50%.
46. a. The graph represents exponential decay.
b. Because f (x) is an exponential function and any real
number can be used as an exponent, the domain is all real
numbers. Because f (x) → 0 as x → ∞ and f (x) → ∞ as
x → −∞, the range is y > 0.
47. Your friend is incorrect. f (x) = 2x does not have a faster
growth rate for x ≥ 0. For instance, the secant line between
(1, 1) and (2, 4) on g(x) = x2 is steeper than the secant line
between (1, 2) and (2, 4) on f (x).
48. Sample answer: The function g(x) = (1 − b)x is an
exponential decay function because 1 − b < 1.
49. The average rate of change over the fi rst 6 years is
6850(1.03)6 − 6850(1.03)0
——— 6 − 0
≈ 8179 − 6850 ——
6 = 221.5.
So, the average rate of change over the fi rst 6 years is about
221.5 people per year.
50. a. f (x + 1)
— f (x)
= abx+1
— abx
= bx+1
— bx
= bx+1−x
= b
b. By part (a), the ratios of consecutive terms must be equal
for the points to be modeled by an exponential function.
f (1)
— f (0)
= 4 — 4 = 1
and
f (2)
— f (1)
= 8 — 4 = 2
The ratios are not equal, so there is no exponential
function of the form f (x) = abx whose graph passes
through the points in the table.
51. a. The decay factor is 0.89 and the percent decrease is 0.11,
or 11%.
b. Eggs Produced by Leghorn
50 100 150 200 250 3000 w
E
90
100
110
120
130
140
150
160
170
180
800
Age (weeks)
Nu
mb
er o
f eg
gs
c. A chicken that is 2.5 years old is 2.5(52) = 130 weeks
old and produces about E = 179.2(0.89)130/52 ≈ 134 eggs
per year.
d. To use years, rather than weeks, replace w
— 52
with y, where
y is the age in years.
52. Use the model V = abt. When the stereo is new, t = 0 and
V = 1300, so 1300 = ab0 = a. After 4 years, V = 275,
so 275 = 1300b4. Solve for b.
275 = 1300b4
0.2115 = b4
0.6782 ≈ b So, the model for the value of stereo is V = 1300(0.6782)t.
Maintaining Mathematical Profi ciency
53. x9 ⋅ x2 = x9 + 2 = x11
54. x4 —
x3 = x4−3 = x
55. 4x ⋅ 6x = 24x1+1 = 24x2
56. ( 4x8 —
2x6 )
4
= (2x8−6)4 = (2x2)4 = 16x8
57. x + 3x — 2 = 4x
— 2 = 2x
58. 6x — 2 + 4x = 3x + 4x = 7x
59. 12x — 4x
+ 5x = 3 + 5x
60. (2x ⋅ 3x5)3 = (6x6)3 = 216x18
6.2 Explorations (p. 303)
1. e ≈ 2.718; Sample answer: Add a few more values to the
sequence shown by following its pattern. Use a graphing
calculator to fi nd the value of each fraction, and add these
values; 3
Copyright © Big Ideas Learning, LLC Algebra 2 297All rights reserved. Worked-Out Solutions
Chapter 6
2. x 101 102 103
( 1 + 1 — x ) x 2.59374 2.70481 2.71692
x 104 105 106
( 1 + 1 — x )
x
2.71815 2.71827 2.71828
Sample answer: e ≈ 2.71828; This approximation has more
decimal points.
3. x −2 −1 0 1 2
y = ex 0.1353 0.3679 1 2.7183 7.3891
y
21−1−2
2
4
6
8
x
The domain is all real numbers and the range is y > 0. It is
an exponential function and any real number can be used as
an exponent. f (x) → 0 as x → −∞ and f (x) → ∞ as x → ∞.
4. The natural base e ≈ 2.71828 is an irrational number that
can be found by approximating
( 1 + 1 — x )
x
or 1 + 1 — 1 + 1
— 1 ⋅ 2
+ 1 —
1 ⋅ 2 ⋅ 3
+ 1 ——
1 ⋅ 2 ⋅ 3 ⋅ 4 + ∙ ∙ ∙∙
5. x −2 −1 0 1 2
y = e−x 7.3891 2.7183 1 0.3679 0.1353
y
21−1−2
2
4
6
8
x
The domain is all real numbers and the range is y > 0. The
graph of y = e−x is a refl ection in the y-axis of the graph of
y = ex. The domain and range of each graph are the same.
6. Sample answer: The natural base e is used in some
population models.
6.2 Monitoring Progress (pp. 304–306)
1. e7 ⋅ e4 = e7+4
= e11
2. 24e8 —
8e5 = 3e8−5
= 3e3
3. (10e−3x)3 = 103(e−3x)3
= 1000e−9x
= 1000 —
e9x
4. Because a = 1 — 2 is positive and r = 1 is positive, the function
is an exponential growth function.
Use a table to graph the function.
x −2 −1 0 1 2
y 0.07 0.18 0.5 1.36 3.69
y
42−2−4
2
4
6
8
x
5. Because a = 4 is positive and r = −1 is negative, the
function is an exponential decay function.
Use a table to graph the function.
x −2 −1 0 1
y 29.56 10.87 4 1.47
y
42−2−4
2
4
8
x
298 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
6. Because a = 2 is positive and r = 2 is positive, the function
is an exponential growth function.
Use a table to graph the function.
x −2 −1 0 1
y 0.04 0.27 2 14.78
y
42−2−4
2
4
6
8
x
7. Understand the Problem You are given the principal
amount and the interest rate compounded continuously. You
are asked to compare the balance after 10 years with the
accounts in Example 3.
Make a Plan Write an equation to calculate the balance
after 10 years when interest is compounded continuously.
Then compare the balance with the accounts in Example 3.
Solve the Problem Let P = 4250 and r = 0.05.
The amount A in an account after t years is given by
A = Pert = 4250e0.05t.
When t = 10, A = 4250e(0.05)(10) = $7007.07.
So, this account has a greater balance than your account in
Example 3, but a lesser balance than your friend’s account in
Example 3.
6.2 Exercises (pp. 307–308)
Vocabulary and Core Concept Check
1. an irrational number that is approximately 2.718281828
2. Because a = 1 — 3 is positive and r = 4 is positive, the function
is an exponential growth function.
Monitoring Progress and Modeling with Mathematics
3. e3 ⋅ e5 = e3+5
= e8
4. e −4 ⋅ e6 = e−4+6
= e2
5. 11e9
— 22e10
= 1 — 2 e9−10
= 1 — 2 e−1
= 1 — 2e
6. 27e7
— 3e4
= 9e7−4
= 9e3
7. (5e7x)4 = 54(e7x)4
= 625e28x
8. (4e−2x)3 = 43(e−2x)3
= 64e−6x
= 64 —
e6x
9. √—
9e6x = √—
9 √—
e6x
= 91/2e6x�2
= 3e3x
10. 3 √—
8e12x = 3 √—
8 3 √—
e12x
= 81�3 e12x�3
= 2e4x
11. ex ⋅ e−6x ⋅ e8 = ex+(−6x)+8
= e−5x+8
12. ex ⋅ e4 ⋅ ex+3 = ex+4+(x+3)
= e2x+7
13. The constant 4 needs to be squared.
(4e3x)2 = 42e(3x)(2)
= 16e6x
14. In the exponent, the −2x should be subtracted from 5x.
e5x
— e−2x = e5x−(−2x)
= e5x+2x
= e7x
15. Because a = 1 is positive and r = 3 is positive, the function
is an exponential growth function.
Use a table to graph the function.
x −2 −1 0 1
y 0.002 0.05 1 20.09
y
42−2−4
2
4
6
8
x
Copyright © Big Ideas Learning, LLC Algebra 2 299All rights reserved. Worked-Out Solutions
Chapter 6
16. Because a = 1 is positive and r = −2 is negative, the
function is an exponential decay function.
Use a table to graph the function.
x −2 −1 0 1
y 54.6 7.39 1 0.14
y
42−2−4
4
6
8
x
17. Because a = 2 is positive and r = −1 is negative, the
function is an exponential decay function.
Use a table to graph the function.
x −2 −1 0 1
y 14.78 5.44 2 0.74
y
42−2−4
2
4
6
8
x
18. Because a = 3 is positive and r = 2 is positive, the function
is an exponential growth function.
Use a table to graph the function.
x −2 −1 0 1
y 0.05 0.41 3 22.17
y
42−2−4
2
4
6
8
x
19. Because a = 0.5 is positive and r = 1 is positive, the
function is an exponential growth function.
Use a table to graph the function.
x −2 −1 0 1 2
y 0.07 0.18 0.5 1.36 3.69
y
42−2−4
2
4
6
8
x
20. Because a = 0.25 is positive and r = −3 is negative, the
function is an exponential decay function.
Use a table to graph the function.
x −2 −1 0 1
y 100.86 5.02 0.25 0.01
y
42−2−4
4
6
8
x
21. Because a = 0.4 is positive and r = −0.25 is negative, the
function is an exponential decay function.
Use a table to graph the function.
x −2 −1 0 1
y 0.66 0.51 0.4 0.31
y
4−4−8
4
8
x
16
12
−12
300 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
22. Because a = 0.6 and r = 0.5 are positive, the function is an
exponential growth function.
Use a table to graph the function.
x −2 −1 0 1
y 0.22 0.36 0.6 0.99
y
42−2−4
4
2
6
8
x
23. The graph is D because the function represents exponential
growth and (0, 1) is a point on the graph of the function.
24. The graph is A because the function represents exponential
decay and (0, 1) is a point on the graph of the function.
25. The graph is B because the function represents exponential
decay and (0, 4) is a point on the graph of the function.
26. The graph is C because the function represents exponential
growth and (0, 0.75) is a point on the graph of the function.
27. y = e−0.25t
= (e−0.25)t
≈ (0.779)t
= (1 − 0.221)t
The percent decrease is about 22.1%.
28. y = e−0.75t
= (e−0.75)t
≈ (0.472)t
= (1 − 0.528)t
The percent decrease is about 52.8%.
29. y = 2e0.4t
= 2(e0.4)t
≈ 2(1.492)t
= 2(1 + 0.492)t
The percent increase is about 49.2%.
30. y = 0.5e0.8t
= 0.5(e0.8)t
≈ 0.5(2.226)t
= 0.5(1 + 1.226)t
The percent increase is about 122.6%.
31. x −2 −1 0 1 2
y 0.02 0.05 0.14 0.37 1
y
42−2−4
4
2
6
8
x
The domain of the function is all real numbers and the range
of the function is y > 0.
32. x −2 −1 0 1 2
y 0.37 1 2.72 7.39 20.09
y
42−2−4
4
6
8
x
The domain of the function is all real numbers and the range
of the function is y > 0.
33. x −2 −1 0 1
y 1.27 1.74 3 6.44
y
42−2−4
8
4
x
16
12
The domain of the function is all real numbers and the range
of the function is y > 1.
Copyright © Big Ideas Learning, LLC Algebra 2 301All rights reserved. Worked-Out Solutions
Chapter 6
34. x −2 −1 0 1
y −4.59 −3.90 −2 3.15
y
42−2−4
8
4
x
12
−6
The domain of the function is all real numbers and the range
of the function is y > −5.
35. Understand the Problem You are given a graph and an
equation that represent account balances. You are asked
to identify the account with the greater principal and the
account with the greater balance after 10 years.
Make a Plan Use the equation to fi nd the principal and
account balance of the house fund after 10 years. Then
compare these values to the graph of the education account.
Solve the Problem The equation H = 3224e 0.05t is of
the form H = Pert, where P = 3224. So, the principal of
the house account is $3224. The balance when t = 10 is
H = 3224e0.05(10) ≈ $5315.48.
Because the graph passes through (0, 4856), the principal of
the education account is $4856. The graph also shows that
the balance is about $7250 when t = 10.
So, the education account has the greater principal and
greater balance after 10 years than the house account.
36. Understand the Problem You are given a graph and an
equation that represent amounts of elements in a sample over
time. You are asked to identify the sample that started with
a greater amount and the sample that has a greater amount
after 10 years.
Make a Plan Use the equation to calculate the starting
amount and remaining amount of tritium after 10 years. Then
compare these values to the graph of sodium-22 decay.
Solve the Problem The equation y = 10e−0.0562t is of the
form y = Pert, where P = 10. So, the starting amount of the
tritium sample is 10 milligrams.
The amount remaining after 10 years is
y = 10e−0.0562(10) ≈ 5.7 milligrams.
Because the graph passes through (0, 15), the starting
amount of sodium-22 in a sample is 15 milligrams. The
graph also shows that the remaining amount after 10 years is
about 1.25 milligrams.
So, sodium-22 started with a greater amount than tritium, but
the tritium sample has a greater amount after 10 years than
the sodium-22 sample.
37. Sample answer: a = 1, b = 2, r = −2, q = −5. So,
f (x) = e−2x and g(x) = 2e−5x are exponential decay
functions, and f (x)
— g(x)
= e−2x
— 2e−5x =
1 —
2 e3x is an exponential growth
function.
38. Let m = n —
r , so n = mr and
r —
n =
1 —
m .
Substituting into A = P ( 1 + r —
n )
nt gives A = P ( 1 +
1 —
m )
mrt
which can be written as A = P [ ( 1 + 1 —
m )
m
] rt
. By defi nition,
( 1 + 1 —
m )
m
approaches e as m approaches +∞. So, the
equation becomes A = Pert.
39. no; The natural base e is an irrational number, so it cannot be
written as the ratio of two integers.
40. Your friend is incorrect. Because y = f (x) is an exponential
function, it does not have an x-intercept.
41. For account 1, P = 2500, n = 4, t = 10, and r = 0.06, so the
balance in 10 years is
A = 2500 ( 1 + 0.06
— 4 )
40
≈ $4535.05.
For account 2, P = 2500, t = 10, and r = 0.04, so the
balance in 10 years is
A = 2500e0.04(10)
≈ $3729.56.
So, you should choose account 1 to obtain the greater
amount in 10 years.
42. a. f (x) increases as x increases, so f (x) approaches +∞ as
x approaches +∞.
b. f (x) approaches −3 as x approaches −∞.
43. a. After 1:00 p.m. the number of bacteria is given by the
function N(t) = 30e0.166t.
b.
200
900
0
c. At 3:45 p.m., t = 2.75 hours.
So, to fi nd the number of cells in the sample at 3:45 p.m.,
substitute 2.75 for t in the equation in part (a) and simplify.
At 3:45 p.m., there are about 47 cells.
Maintaining Mathematical Profi ciency
44. 0.006 = 6 × 10−3
45. 5000 = 5 × 103
46. 26,000,000 = 2.6 × 107
47. 0.000000047 = 4.7 × 10−8
302 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
48. y = 3x + 5
x = 3y + 5
x − 5 = 3y
x − 5
— 3 = y
So, the inverse function is y = x − 5
— 3 .
y
2−4−6
2
x
−4
−6
y = 3x + 5
y = x − 5
3
49. y = x2 − 1, x ≤ 0
x = y2 − 1, y ≤ 0
x + 1 = y2
± √—
x + 1 = y
Because x ≤ 0 in the original equation, the range of the inverse is y ≤ 0. So, choose the negative root.
So, the inverse function is y = − √—
x + 1 .
y
4 62−2
4
6
2
x
−2
y = x2 − 1, x ≤ 0
y = −√x + 1——
50. y = √—
x + 6
x = √—
y + 6
x2 = y + 6
x2 − 6 = y
Because y ≥ 0 in the original equation, the domain of the
inverse is x ≥ 0.
So, the inverse
function is
y = x2 − 6, x ≥ 0.
51. y = x3 − 2
x = y3 − 2
x + 2 = y3
3 √—
x + 2 = y
So, the inverse function is y = 3 √—
x + 2 .
y
4 62−4−6
4
6
2
x
−4
−6
y = x3 − 2
y = √3x + 2——
6.3 Explorations (p. 309)
1. a. The value of x is 3 because 23 = 8; 3 = log2 8.
b. The value of x is 2 because 32 = 9; 2 = log3 9.
c. The value of x is 1 —
2 because 41�2 = 2;
1 —
2 = log4 2.
d. The value of x is 0 because 50 = 1; 0 = log5 1.
e. The value of x is −1 because 5−1 = 1 —
5 ; −1 = log5
1 —
5 .
f. The value of x is 2 —
3 because 82�3 = 4;
2 —
3 = log8 4.
2. a. x −2 −1 0 1 2
f (x) = 2x 2−2 = 1 —
4 2−1 =
1 —
2 20 = 1 21 = 2 22 = 4
The functions f and g are inverses, so the x- and
y-coordinates can be switched to create the table for g.
x 1 —
4
1 —
2 1 2 4
g (x) = log2 x −2 −1 0 1 2
y
42−2
2
4
x
−2
g
f
y
4 6 82−4 −2−6
4
6
2
x
−2
−4
−6
y = x2 − 6, x ≥ 0
y = √x + 6——
Copyright © Big Ideas Learning, LLC Algebra 2 303All rights reserved. Worked-Out Solutions
Chapter 6
b.
x −2 −1 0 1 2
f (x) = 10 x 10−2 = 1 —
100 10−1 =
1 —
10 100 = 1 101 = 10 102 = 100
The functions f and g are inverses, so the x- and
y-coordinates can be switched to create the table for g.
x 1 —
100
1 —
10 1 10 100
g (x) = log10 x −2 −1 0 1 2
y
4 6 8 102−2
2
4
6
8
x
−2
10
g
f
3. The graph of g(x) = logb x is a refl ection in the line y = x of the graph of f (x) = b x. So, the domain of g is x > 0, the
range is all real numbers, the x-intercept is (1, 0), and the
asymptote is x = 0.
4. The graph of the logarithmic function has a domain of x > 0
and a range of all real numbers. The graph has an asymptote
at x = 0.
5. You can use the refl ection of the exponential function in the
y = x line to graph the logarithmic function.
6.3 Monitoring Progress (pp. 311–313)
1. log3 81 = 4, so 34 = 81.
2. log7 7 = 1, so 71 = 7.
3. log14 1 = 0, so 140 = 1.
4. log1�2 32 = −5, so ( 1 — 2 ) −5
= 32.
5. 72 = 49, so log7 49 = 2.
6. 500 = 1, so log50 1 = 0.
7. 4−1 = 1 — 4 , so log4
1 — 4 = −1.
8. 2561�8 = 2, so log256 2 = 1 — 8 .
9. log2 32 = 5
10. log27 3 = 1 — 3
11. log 12 ≈ 1.079
12. ln 0.75 ≈ −0.288
13. 8 log8 x = x
14. log7 7−3x = −3x
15. log2 64x = log2 (26)x
= log2 26x
= 6x
16. eln 20 = 20
17. y = 4x
x = 4y
log4 x = log4 4y
log4 x = y The inverse of y = 4x is y = log4 x.
18. y = ln(x − 5)
x = ln( y − 5)
ex = y − 5 ex + 5 = y The inverse of y = ln(x − 5) is y = ex + 5.
19. Step 1 Find the inverse of y. From the defi nition of
logarithm, the inverse of y = log2 x is y = 2x.
Step 2 Make a table of values for y = 2x.
x −2 1 0 1 2
y 1 —
4
1 —
2 1 2 4
Step 3 Plot the points from the table and connect them with
a smooth curve.
Step 4 Because y = log2 x and y = 2x are inverse functions,
the graph of y = log2 x is obtained by refl ecting the
graph of y = 2x in the line y = x. To do this, reverse
the coordinates of the points of y = 2x and plot these
new points to obtain the graph of y = log2 x.
y
4 6 8 102
2
4
6
8
x
−2
304 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
20. Step 1 Find the inverse of f. From the defi nition of
logarithm, the inverse of f (x) = log5 x is g(x) = 5x.
Step 2 Make a table of values of g(x) = 5x.
x −2 −1 0 1 2
g(x) 1 —
25
1 —
5 1 5 25
Step 3 Plot the points from the table and connect them with
a smooth curve.
Step 4 Because f (x) = log5 x and g(x) = 5x are inverse
functions, the graph of f is obtained by refl ecting
the graph of g in the line y = x. To do this, reverse
the coordinates of the points of g and plot these new
points to obtain the graph of f.
y
4 62
2
4
x
−2
21. Step 1 Find the inverse of y = log1�2 x. From the defi nition
of logarithm, the inverse of y = log1�2 x is y = ( 1 — 2 )
x
.
Step 2 Make a table of values of y = ( 1 — 2 )
x
.
x −2 −1 0 1 2
y 4 2 1 1 —
2
1 —
4
Step 3 Plot the points from the table and connect them with
a smooth curve.
Step 4 Because y = log1�2 x and y = ( 1 — 2 )
x
are inverse
functions, the graph of y = log1�2 x is obtained
by refl ecting the graph of y = ( 1 — 2 )
x
in the line
y = x. To do this, reverse the coordinates of the
points of y = ( 1 — 2 )
x
and plot these points to obtain the
graph of y = log1�2 x.
y
4 62
2
4
x
−2
6.3 Exercises (pp. 314 –316)
Vocabulary and Core Concept Check
1. A logarithm with base 10 is called a common logarithm.
2. The expression log3 9 is read as “log base 3 of 9.”
3. The functions y = 7x and y = log7 x are inverse functions.
4. “Evaluate 42 ” is different than the others, the answer of
“Evaluate 42 ” is 16 and the other answer is 2.
Monitoring Progress and Modeling with Mathematics
5. log3 9 = 2, so 32 = 9.
6. log4 4 = 1, so 41 = 4.
7. log6 1 = 0, so 60 = 1.
8. log7 343 = 3, so 73 = 343.
9. log1�2 16 = −4, so ( 1 — 2 )
−4
= 16.
10. log3 1 —
3 = −1, so 3−1 = 1 —
3 .
11. 62 = 36, so log6 36 = 2.
12. 120 = 1, so log12 1 = 0.
13 16−1 = 1 — 16
, so log16 1 —
16 = −1.
14. 5−2 = 1 — 25
, so log5 1 —
25 = −2.
15. 1252�3 = 25, so log125 25 = 2 — 3 .
16. 491�2 = 7, so log49 7 = 1 — 2 .
17. log3 81 = log3 34 = 4
18. log7 49 = log7 72 = 2
19. log3 3 = 1
20. log1�2 1 = 0
21. log5 1 —
625 = log5 5
−4 = −4
22. log8 1 —
512 = log8 8
−3 = −3
23. log4 0.25 = log4 1 —
4 = log4 4−1 = −1
24. log10 0.001 = log10 1 —
1000 = log10 10−3 = −3
25. log5 23 ≈ 1.95, log6 38 ≈ 2.03, log7 8 ≈ 1.07,
log8 10 ≈ 3.32 so, log7 8 < log5 23 < log6 38 < log2 10.
26. The expression log2 (−1) is not defi ned because there is no
exponent for 2 that produces −1. The expression log1 1 is
not defi ned because the logarithm is not defi ned for a base
of 1.
27. log 6 ≈ 0.778
28. ln 12 ≈ 2.485
29. ln 1 —
3 ≈ −1.099
30. log 2 —
7 ≈ −0.544
31. 3 ln 0.5 ≈ −2.079
Copyright © Big Ideas Learning, LLC Algebra 2 305All rights reserved. Worked-Out Solutions
Chapter 6
32. log 0.6 + 1 ≈ 0.778
33. When p = 57,000,
h = −8005 ln 57,000
— 101,300
≈ 4603.
So, when the air pressure is 57,000 pascals the height is about 4603 meters.
34. a. pH = −log 10−8 = −(−8) = 8 The pH value of baking soda is 8.
b. pH = −log 10−3 = −(−3) = 3 The pH value of vinegar is 3.
35. 7 log7 x = x
36. 3 log3 5x = 5x
37. eln 4 = 4
38. 10log 15 = 15
39. log3 32x = 2x
40. ln ex+1 = x + 1
41. The error is in the placement of the values.
log4 1 —
64 = −3
42. The error is in rewriting 64x.
log4 64x = log4 43x = 3x
43. y = 0.3x
x = 0.3y
log0.3 x = y The inverse of y = 0.3x is y = log0.3 x.
44. y = 11x
x = 11y
log11 x = y The inverse of y = 11x is y = log11 x.
45. y = log2 x
x = log2 y
2x = y The inverse of y = log2 x is y = 2x.
46. y = log1�5 x x = log1�5 y
( 1 — 5 )
x
= y
The inverse of y = log1�5 x is y = ( 1 — 5 )
x
.
47. y = ln(x − 1)
x = ln( y − 1)
ex = y − 1 ex + 1 = y The inverse of y = ln(x − 1) is y = ex + 1.
48. y = ln 2x
x = ln 2y
ex = 2y
1 — 2 ex = y
The inverse of y = ln 2x is y = 1 — 2 ex.
49. y = e3x
x = e3y
ln x = 3y
1 —
3 ln x = y
The inverse of y = e3x is y = 1 — 3 ln x.
50. y = ex−4
x = ey−4
ln x = y − 4 ln x + 4 = y The inverse of y = ex−4 is y = ln x + 4.
51. y = 5x − 9 x = 5y − 9 x + 9 = 5y
log5(x + 9) = y The inverse of y = 5x − 9 is y = log5(x + 9).
52. y = 13 + log x
x = 13 + log y
x − 13 = log y
10x−13 = y The inverse of y = 13 + log x is y = 10x−13.
53. a. When d = 220,
s = 93 log 220 + 65
≈ 93(2.34) + 65
= 282.84.
The wind speed was about 283 miles per hour.
b. s = 93 log d + 65
s − 65 = 93 log d
s − 65 —
93 = log d
10 (s−65)/93 = d
The inverse is d = 10(s−65)/93 and represents the distance d
a tornado travels given the wind speed s.
306 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
54. a. When E = 2.24 × 1028,
M = 2 — 3 log(2.24 × 1028) − 9.9
≈ 2 — 3 (28.35) − 9.9
≈ 9.00.
So, when the energy released is 2.24 × 1028 ergs, the
magnitude of earthquake is about 9.
b. M = 2 — 3 log E − 9.9
M + 9.9 = 2 — 3 log E
3 — 2 (M + 9.9) = log E
10 3/2 (M+9.9) = E
The inverse is E = 10 3/2 (M+9.9) and represents the energy
E released from an earthquake of magnitude M.
55. Step 1 Find the inverse of y = log4 x. From the defi nition of
logarithm, the inverse of y = log4 x is y = 4x.
Step 2 Make a table of values for y = 4x.
x −2 −1 0 1 2
y 1 —
16
1 —
4 1 4 16
Step 3 Plot the points from the table and connect them with
a smooth curve.
Step 4 Because y = log4 x and y = 4x are inverse functions,
the graph of f is obtained by refl ecting the graph
of y = 4x in the line y = x. To do this, reverse the
coordinates of the points of y = 4x and plot these
new points to obtain the graph of y = 4x.
y
4 62
2
4
x
−2
56. Step 1 Find the inverse of y = log6 x. From the defi nition of
logarithm, the inverse of y = log6 x is y = 6x.
Step 2 Make a table of values of y = 6x.
x −2 −1 0 1 2
y 1 —
36
1 —
6 1 6 36
Step 3 Plot the points from the table and connect them with
a smooth curve.
Step 4 Because y = log6 x and y = 6x are inverse functions,
the graph of y = log6 x is obtained by refl ecting the
graph of y = 6x in the line y = x. To do this, reverse
the coordinates of the points of y = 6x and plot these
new points to obtain the graph of y = log6 x.
y
4 62
2
4
x
−2
57. Step 1 Find the inverse of y = log1/3 x. From the defi nition
of logarithm, the inverse of y = log1/3 x is y = ( 1 — 3 )
x
.
Step 2 Make a table of values of y = ( 1 — 3 )
x
.
x −2 −1 0 1 2
y 9 3 1 1 —
3
1 —
9
Step 3 Plot the points from the table and connect them with
a smooth curve.
Step 4 Because y = log1�3 x and y = ( 1 — 3 )
x
are inverse
functions, the graph of y = log1�3 x is obtained by
refl ecting the graph of y = ( 1 — 3 )
x
in the line y = x.
To do this, reverse the coordinates of the points of
y = ( 1 — 3 )
x
and plot these new points to obtain the
graph of y = log1�3 x.
y
4 6 8 10
2
4
6
x
−2
−4
Copyright © Big Ideas Learning, LLC Algebra 2 307All rights reserved. Worked-Out Solutions
Chapter 6
58. Step 1 Find the inverse of y = log1�4 x. From the defi nition
of logarithm, the inverse of y = log1�4 x is y = ( 1 — 4 )
x
.
Step 2 Make a table of values for y = ( 1 — 4 )
x
.
x −2 −1 0 1 2
y 1 —
16
1 —
4 1 4 16
Step 3 Plot the points from the table and connect them with
a smooth curve.
Step 4 Because y = log1�4 x and y = ( 1 — 4 )
x
are inverse
functions, the graph of y = log1�4 x is obtained by
refl ecting the graph of y = ( 1 — 4 )
x
in the line y = x.
To do this, reverse the coordinates of the points of
y = ( 1 — 4 )
x
and plot these new points to obtain the
graph of y = log1�4 x.
y
4 6
2
4
x
−2
59. Step 1 Find the inverse of y = log2 x − 1. From the
defi nition of logarithm, the inverse of y = log2 x − 1
is y = 2x+1.
Step 2 Make a table of values for y = 2x+1.
x −2 −1 0 1 2
y 1 —
2 1 2 4 8
Step 3 Plot the points from the table and connect them with
a smooth curve.
Step 4 Because y = log2 x − 1 and y = 2x+1 are inverse
functions, the graph of y = log2 x − 1 is obtained
by refl ecting the graph of y = 2x+1 in the line
y = x. To do this, reverse the coordinates of the
points of y = 2x+1 and plot these new points to
obtain the graph of y = log2 x − 1.
y
42 6 8
2
4
x
−2
−4
60. Step 1 Find the inverse of y = log3(x + 2). From the
defi nition of logarithm, the inverse of y = log3(x + 2)
is y = 2x − 2.
Step 2 Make a table of values for y = 2x − 2.
x −2 −1 0 1 2
y − 7 — 4 − 3 —
2 −1 0 2
Step 3 Plot the points from the table and connect them with
a smooth curve.
Step 4 Because y = log3(x + 2) and y = 2x − 2 are inverse
functions, the graph of y = log3(x + 2) is obtained
by refl ecting the graph of y = 2x − 2 in the line
y = x. To do this, reverse the coordinates of the
points of y = 2x − 2 and plot these new points to
obtain the graph of y = log3(x + 2).
y
42 6 8
2
4
x
−2
−4
−2
61. y
42
2
4
x
−2
−2
The domain of the function is
x > −2 and the asymptote is
x = −2.
62. y
4 6
2
4
x
−2
The domain of the function is
x > 0 and the asymptote is x = 0.
63. y
42
2
4
x
−2
−4
−2−4
The domain of the function
is x < 0 and the asymptote
is x = 0.
308 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
64. y
4 62
2
4
x
−2
The domain of the function is
x > 0 and the asymptote is
x = 0.
65. Your friend is incorrect because a logarithmic function that
has been translated, stretched or refl ected in the x-axis may
not pass through (1, 0).
66. a. log6 10 − log6 1 —— 10 − 1
≈ 0.14
b. log3�5 10 − log3�5 1 —— 10 − 1
≈ −0.50
c. f (10) − f (1) ——
10 − 1 ≈ −1 − 0
— 10 − 1 = −
1 —
10 = −0.10
d. g(10) − g(1) ——
10 − 1 ≈ 8 − 0
— 10 − 1 = 0.80
The order from least to greatest is b, c, a, and d.
67. a.
2000
180
0
b. Using the trace feature, when l = 120, w ≈ 281, so the
weight is about 281 pounds.
c. The zero of the function is about 3.4. This x-value does
not make sense because an alligator cannot have a length
of 0 inches.
68. a. As x → −∞, f (x) → 0. As x → ∞, f (x) → ∞ and
g(x) → ∞.
b. The functions are inverses because they are refl ections
of each other in the line y = x.
c. The logarithmic function passes through (6, 1), so the
base is 6 and g(x) = log6 x. Because f is the inverse of g,
f (x) = 6x, so the base of the exponential function is 6.
69. a.
35,0000
20
200
b. Using the trace feature, when A = 30,000, s = 15. So,
there are about 15 fi sh species.
c. Using the trace feature, when s ≈ 6, A ≈ 3917. So, the
area is about 3917 square meters.
d. When the size increases, the number of species of fi sh
increases. Sample answer: This makes sense because
more species can coexist in a larger space.
70. Because the range of a logarithmic function is all real
numbers, −4 is an output of any logarithmic function.
Sample answer: f (x) = log(x) − 5 y
42 6 8 10
2
4
x
−2
−4
−6
71. a. log135 25 = log 5 3 52 = 2 —
3
b. log8 32 = log 2 3 25 = 5 —
3
c. log27 81 = log 3 3 34 = 4 —
3
d. log4 128 = log 2 2 27 = 7 —
2
Maintaining Mathematical Profi ciency
72. g(x) = −f (x) = − 3 √—
x
73. g(x) = f ( 1 — 2 x ) =
3 √—
1 —
2 x
74. g(x) = f (−x) + 3 = 3 √—
−x + 3
75. g(x) = f (x + 2) = 3 √—
x + 2
76. The parent function is the constant function g(x) = 1; f is a
translation 3 units down of the graph of g.
77. The parent function is the quadratic function g(x) = x2; f is a
translation 2 units left and 1 unit down of the graph of g.
78. The parent function is the absolute value function g(x) = ∣ x ∣ ; f is a refl ection in the x-axis followed by a translation 2 units
up and 1 unit right of the graph of g.
6.4 Explorations (p. 317)
1. a. F: The graph shows the parent function f (x) = e x translated 2 units left and 3 units down.
b. C: The graph shows the parent function f (x) = e x refl ected
in the x-axis then translated 2 units left and 1 unit up.
c. A: The graph shows the parent function f (x) = e x translated 2 units right and 1 unit down.
d. E: The graph shows the parent function f (x) = ln x
translated 2 units left.
e. D: The graph shows the parent function f (x) = ln x
translated 2 units up.
f. B: The graph shows the parent function f (x) = ln x
refl ected in the y-axis and translated 2 units up.
Copyright © Big Ideas Learning, LLC Algebra 2 309All rights reserved. Worked-Out Solutions
Chapter 6
2. a. domain: all real numbers, range: y > −3, asymptote:
y = −3; The graph exists for all values of x and there is
an asymptote at y = −3.
b. domain: all real numbers, range: y < 1, asymptote:
y = 1; The graph exists for all values of x and there is an
asymptote at y = 1.
c. domain: all real numbers, range: y > −1, asymptote:
y = −1; The graph exists for all values of x and there is
an asymptote at y = −1.
d. domain: x > −2, range: all real numbers, asymptote:
x = −2; There is an asymptote at x = −2 and the graph
exists for all values of y.
e. domain: x > 0, range: all real numbers, asymptote: x = 0;
There is an asymptote at x = 0 and the graph exists for all
values of y.
f. domain: x < 0, range: all real numbers, asymptote: x = 0;
There is an asymptote at x = 0 and the graph exists for all
values of y.
3. The graphs of exponential and logarithmic functions can be
translated using the same rules as all other functions.
4. a. g(x) = e x+2 − 3 y = e x+2 − 3 x = e y+2 − 3 x + 3 = e y+2
ln(x + 3) = y + 2 ln (x + 3) − 2 = y The inverse of g(x) = e x+2 − 3 is y = ln(x + 3) − 2
5
−5
−7
3
b. g(x) = −e x+2 + 1 y = −e x+2 + 1 x = −e y+2 + 1 x − 1 = −e y+2
−x + 1 = e y+2
ln(−x + 1) = y + 2 ln(−x + 1) − 2 = y The inverse of g(x) = −e x+2 + 1 is y = ln(−x + 1) − 2.
4
−6
−8
2
c. g(x) = e x−2 − 1 y = e x−2 − 1 x = e y−2 − 1 x + 1 = e y−2
ln(x + 1) = y − 2 ln(x + 1) + 2 = y The inverse of g(x) = e x−2 − 1 is y = ln(x + 1) + 2.
6
−3
−6
5
d. g(x) = ln(x + 2)
y = ln(x + 2)
x = ln(y + 2)
e x = y + 2 e x − 2 = y The inverse of g(x) = ln(x + 2) is y = e x − 2.
6
−4
−6
4
e. g(x) = 2 + ln x
y = 2 + ln x
x = 2 + ln y
x − 2 = ln y
e x−2 = y The inverse of g(x) = 2 + ln x is y = e x−2.
6
−3
−6
5
f. g(x) = 2 + ln(−x)
y = 2 + ln(−x)
x = 2 + ln(−y)
x − 2 = ln(−y)
e x−2 = −y
−e x−2 = y The inverse of g(x) = 2 + ln(−x) is y = −e x−2.
6
−4
−6
4
310 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
6.4 Monitoring Progress (pp. 318–321)
1. Notice that the function is of the form g(x) = 2 x−h + k,
where h = 3 and k = 1. So, the graph of g is a translation
3 units right and 1 unit up of the graph of f.
y
42 6
2
4
6
x−2
g
f
2. Notice that the function is of the form g(x) = e−x + k, where k = −5. So, the graph of g is a translation 5 units,
down of the graph of f.
y
42
4
x−2−4
−4
−6
−2g
f
3. Notice that the function is of the form g(x) = 0.4−ax, where
a = 2. So, the graph of g is a refl ection in the y-axis and a
horizontal shrink by a factor of 1 —
2 of the graph of f.
y
42−2−4
4
6
8
x
g
f
4. Notice that the function is of the form g(x) = −ex−h, where
h = −6. So, the graph of g is a translation 6 units left
followed by a refl ection in the x-axis of the graph of f.
y
2−2−4−6−8
4
2
6
x
−4
−6
−2g
f
5. Notice that the function is of the form g(x) = −a log2 x,
where a = 3. So, the graph of g is a vertical stretch by a
factor of 3 and a refl ection in the x-axis of the graph of f.
y
2 4 6 8 10
4
2
x
−4
−6
−8
−2
g
f
6. Notice that the function is of the form
g(x) = log1/4 (ax) + k, where a = 4 and k = −5.
So, the graph of g is a horizontal shrink by a factor of 1 —
4
followed by a translation 5 units down of the graph of f.
y
4 6 8 10
2
x
−4
−6
−8
−2
g
f
7. Step 1 First write a function h that represents the horizontal
stretch.
h(x) = f ( 1 — 3 x )
= e −1/3x
Step 2 Then write a function g that represents the
translation.
g(x) = h(x) + 2 = e −1/3x + 2 The transformed function is g(x) = e −1/3x + 2.
8. Step 1 First write a function h that represents the refl ection.
h(x) = f (−x)
= log(−x)
Step 2 Then write a function g that represents the translation.
g(x) = h(x + 4)
= log [ −(x + 4) ] = log(−x − 4)
The transformed function is g(x) = log(−x − 4).
Copyright © Big Ideas Learning, LLC Algebra 2 311All rights reserved. Worked-Out Solutions
Chapter 6
6.4 Exercises (pp. 322–324)
Vocabulary and Core Concept Check
1. Positive values of a vertically stretch (a > 1) or shrink
(a < 1) the graph of f, h translates the graph of f left (h < 0)
or right (h > 0), and k translates the graph of f up (k > 0) or
down (k < 0). When a is negative, the graph of f is refl ected
in the x-axis.
2. The graph of g(x) = log4(−x) is a refl ection in the y-axis of
the graph of f (x) = log4 x.
Monitoring Progress and Modeling with Mathematics
3. The graph is C because the parent function is translated left
and down.
4. The graph is D because the parent function is translated left
and up.
5. The graph is A because the parent function is translated right
and down.
6. The graph is B because the parent function is translated right
and up.
7. Notice that the function is of the form g(x) = 3x + k.
Because k = 5, the graph of g is a translation 5 units up of
the graph of f.
y
42
4
8
x−2−4
16
12 g f
8. Notice that the function is of the form g(x) = 4x + k. Rewrite
the function to identify k.
g(x) = 4x + (−8)
Because k = −8, the graph of g is a transformation 8 units
down of the graph of f.
y
42
4
8
x−2−4
−4
g
f
9. Notice that the function is of the form g(x) = ex + k. Rewrite
the function to identify k.
g(x) = ex + (−1)
Because k = −1, the graph of g is a translation 1 unit down
of the graph of f.
y
42−2−4
4
2
6
8
x
g
f
10. Notice that the function is of the form g(x) = ex + k.
g(x) = ex + 4
Because k = 4, the graph of g is a translation 4 units up of
the graph of f.
y
42
4
8
x−2−4
16
12 g
f
11. Notice that the function is of the form g(x) = 2x−h. Rewrite
the function to identify h.
g(x) = 2x−7
Because h = 7, the graph of g is a translation 7 units right of
the graph of f.
y
8 124
2
4
6
8
x−4
g
f
312 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
12. Notice that the function is of the form g(x) = 5x−h. Rewrite
the function to identify h.
g(x) = 5x−(−1)
Because h = −1, the graph of g is a translation 1 unit left of
the graph of f.
y
42−2−4
4
6
x
g
f
13. Notice that the function is of the form g(x) = e−x + k.
g(x) = e−x + 6
Because k = 6, the graph of g is a translation 6 units up of
the graph of f.
y
42−2−4
4
6
x
10
g
f
14. Notice that the function is of the form g(x) = e−x + k.
Rewrite the function to identify k.
g(x) = e−x + (−9)
Because k = −9, the graph of g is a translation 9 units down
of the graph of f.
y
42
4
8
x−4
−4
g
f
15. Notice that the function is of the form g(x) = ( 1 — 4 ) x−h
+ k.
g(x) = ( 1 — 4 ) x−3
+ 12
Because h = 3 and k = 12, the graph of g is a translation
3 units right and 12 units up of the graph of f.
y
42 x−4 −2
20
40
60
80
−20
g
f
16. Notice that the function is of the form g(x) = ( 1 — 3 )
x−h
+ k.
Rewrite the function to identify h and k.
g(x) = ( 1 — 3 )
x−(−2)
+ ( − 2 — 3 )
Because h = −2 and k = − 2 — 3 , the graph of g is a translation
2 units left and 2 —
3 unit down of the graph of f.
y
4
6
8
x
−4 −2
2 4
g
f
17. Notice that the function is of the form g(x) = eax, where
a = 2. So, the graph of g is a horizontal shrink by a factor
of 1 —
2 of the graph of f.
42−2−4
4
2
6
8
x
y
g
f
18. Notice that the function is of the
42−2−4
4
2
6
8
x
y
g
f
form g(x) = aex, where a = 4 — 3 .
So, the graph of g is a vertical
stretch by a factor of 4 —
3 of the
graph of f.
Copyright © Big Ideas Learning, LLC Algebra 2 313All rights reserved. Worked-Out Solutions
Chapter 6
19. Notice that the function is of the form g(x) = −2x−h, where
h = 3. So, the graph of g is a refl ection in the x-axis followed
by a translation 3 units right of the graph of f.
y
4 6−2−4
4
2
6
x
−4
−6
−2g
f
20. Notice that the function is of the form g(x) = 4ax−h, where
a = 0.5 and h = 5. So, the graph of g is a horizontal stretch
by a factor of 2 followed by a translation 5 units right of the
graph of f.
y
84 12−4−8
4
2
6
8
x
10
g
f
21. Notice that the function is of the form g(x) = ae−bx, where
a = 3 and b = 6. So, the graph of g is a horizontal shrink by a
factor of 1 —
6 and a vertical stretch by a factor of 3 of the graph of f.
42−2−4
4
6
8
x
y
gf
22. Notice that the function is of the form g(x) = e−ax + k,
where a = 5 and k = 2. So, the graph of g is a horizontal
shrink by a factor of 1 —
5 followed by a translation 2 units up of
the graph of f.
42−2−4
4
x
y
gf
23. Notice that the function is of the form g(x) = a ( 1 — 2 ) x−h
+ k,
where a = 6, h = −5, and k = −2. So, the graph of g is a
vertical stretch by a factor of 6 and a translation 5 units left
and 2 units down of the graph of f.
y
4
2
6
8
x−4−6 −2 2 4
gf
24. Notice that the function is of the form g(x) = − ( 3 — 4 ) x−h
+ k,
where h = 7 and k = 1. So, the graph of g is a refl ection in
the x-axis followed by a translation 7 units right and 1 unit
up of the graph of f.
y
8
4
x−8 −4 4−12 12
12
−4
−8
g
f
25. The error is that the parent function was translated left
instead of up.
y
42−2−4
2
6
8
x
10
314 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
26. The error is in the refl ection. The graph of the parent
function f(x) = 3x should be refl ected in the y-axis not the
x-axis.
y
42−2−4
4
6
8
x
27. Notice that the function is of the form g(x) = a log4 x + k, where a = 3 and k = −5. So, the graph of g is a vertical
stretch by a factor of 3 followed by a translation 5 units
down of the graph of f.
y
x2 4 6 10
−2
−4
−6
−8
2
g
f
28. Notice that the function is of the form g(x) = log1�3 (−x) + k,
where k = 6. So, the graph of g is a refl ection in the y-axis
and a translation 6 units up of the graph of f.
y
4
2
x−2−4−6 4 6
−2
6
8
g
f
29. Notice that the function is the form g(x) = −log1�5(x − h),
where h = 7. So, the graph of g is a refl ection in the x-axis
followed by a translation 7 units right of the graph of f.
y
4
2
x44 8 12 16
−2
6
g
f
30. Notice that the function is of the form
g(x) = log2(x − h) + k, where h = −2 and k = −3. So, the
graph of g is a translation 2 units left and 3 units down of the
graph of f.
y
6 82
2
x−2
−4
−6
g
f
31. The graph is A because it is a translation to the right of
the graph of f.
32. The graph is D because it is a translation to the left of
the graph of f.
33. The graph is C because it is a vertical stretch of the
graph of f.
34. The graph is B because it is a horizontal shrink of the
graph of f.
35. Step 1 First write a function h that represents the
translation of f.
h(x) = f (x) − 2 = 5x − 2 Step 2 Then write a function g that represents the refl ection
of h.
g(x) = h(−x)
= 5−x − 2 The transformed function is g(x) = 5−x − 2.
36. Step 1 First write a function h that represents the
refl ection of f.
h(x) = −f (x)
= − ( 2 — 3 ) x
Step 2 Then write a function g that represents the vertical
stretch and translation of h.
g(x) = 6h(x + 4)
= −6 ( 2 — 3 )
x+4
The transformed function is g(x) = −6 ( 2 — 3 )
x+4
.
Copyright © Big Ideas Learning, LLC Algebra 2 315All rights reserved. Worked-Out Solutions
Chapter 6
37. Step 1 First write a function h that represents the
horizontal shrink of f.
h(x) = f (2x) = e2x
Step 2 Then write a function g that represents the translation
of h.
g(x) = h(x) + 5 = e2x + 5
The transformed function is g(x) = e2x + 5.
38. Step 1 First write a function h that represents the
translation of f.
h(x) = f (x − 4) − 1 = e−x+4 − 1 Step 2 Then write a function g that represents the vertical
shrink of h.
g(x) = 1 — 3 h(x)
= 1 — 3 (e−x+4 − 1)
= 1 —
3 e−x+ 4 −
1 —
3
The transformed function is g(x) = 1 — 3 e−x+4 − 1 —
3 .
39. Step 1 First write a function h that represents a vertical
stretch of f.
h(x) = 6 f (x)
= 6 log 6 x
Step 2 Then write a function g that represents a translation
of h.
g(x) = h(x) − 5 = 6 log 6 x − 5 The transformed function is g(x) = 6 log 6
x − 5.
40. Step 1 First write a function h that represents a
refl ection of f.
h(x) = −f (x)
= −log5 x
Step 2 Then write a function g that represents a translation
of g.
g(x) = h(x + 9)
= −log5(x + 9)
The transformed function is g(x) = −log5(x + 9).
41. Step 1 First write a function h that represents a
translation of f.
h(x) = f (x + 3) + 2 = log1�2 (x + 3) + 2 Step 2 Then write a function g that represents a refl ection of h.
g(x) = h(−x)
= log1�2 (−x + 3) + 2 The transformed function is g(x) = log1�2 (−x + 3) + 2.
42. Step 1 First write a function h that represents a
translation of f.
h(x) = f (x − 3) + 1 = ln(x − 3) + 1 Step 2 Then write a function g that represents a horizontal
stretch of h.
g(x) = h ( 1 — 8 x )
= ln ( 1 — 8 x − 3 ) + 1
The transformed function is g(x) = ln ( 1 — 8 x − 3 ) + 1.
43. h(x) = −f (x) Multiply the output by − 1.
= −log7 x Substitute log7 x for f (x).
g(x) = h(x) − 6 Subtract 6 from the output.
= −log7 x − 6 Substitute − log7 x for h(x).
44. h(x) = 4 f (x) Multiply the output by 4.
= 4 ⋅ 8x Substitute 8x for f (x).
g(x) = h(x + 3) + 1 Add 3 to the input and 1 to the output.
= 4 ⋅ 8x+3 + 1 Replace x with x + 3 in h(x) and add 1
to the output.
45. The graph of g is a translation 4 units up of the graph of f. The asymptote is y = 4.
46. The graph of g is a translation 9 units right of the graph of f. The asymptote is y = 0.
47. The graph of g is a translation 6 units left of the graph of f. The asymptote is x = −6.
48. The graph of g is a translation 13 units up of the graph
of f. The asymptote is x = 0.
49. The transformation of f is a vertical shrink by a factor of
0.118 followed by a translation 0.159 unit up is S.
Sand particle Diameter (mm), d S
Fine sand 0.125 0.052
Medium sand 0.25 0.088
Coarse sand 0.5 0.123
Very coarse sand 1 0.159
50 a. The transformation is a refl ection in the y-axis.
b. no; The result will still be a refl ection in the y-axis.
51. Your friend is correct. Sample answer: A vertical
translation will result in graphs that do not intersect.
52. yes; ln x and ex are inverses, so they are refl ections of
each other in the line y = x.
316 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
53. a. never; A vertical translation does not affect a vertical
asymptote.
b. always; A vertical translation does affect a horizontal
asymptote.
c. always; A horizontal shrink is obtained by multiplying x
by a constant, which does not change the domain.
d. sometimes; Because there is a vertical translation, it is
possible for the graph to have an x-intercept.
54. a. The domain is t ≥ 0 and the range is 0 ≤ P ≤ 100.
b. P = 100(0.99997)12,000
≈ 69.77
There are about 69.77 grams left after 12,000 years.
c. The transformation would be a vertical stretch by a factor
of 55
— 100
= 5.5.
d. The transformation does not affect the domain, but
changes the range to 0 ≤ P ≤ 550.
55. The function h is a translation 2 units left of f1 and h is a
refl ection in the y-axis followed by a translation 2 units left
of g. Rewrite h as h(x) = f (x + 2) and as
h(x) = g [ −(x + 2) ] .
56. Sample answer: y = 3 ⋅ 2x + 2
Maintaining Mathematical Profi ciency
57. (fg)(x) = f (x)g(x)
= x4 ⋅ x2
= x6
When x = 3, the value of the product is
(fg)(3) = 36 = 729.
58. ( f — g ) (x) = f (x)
— g(x)
= 4x6
— 2x3
= 2x3
When x = 5, the value of the quotient is
( f — g ) (5) = 2(5)3 = 250.
59. (f + g)(x) = f (x) + g(x)
= 6x3 + 8x3
= 14x3
When x = 2, the value of the sum is
(f + g)(2) = 14(2)3 = 112.
60. (f − g)(x) = f (x) − g(x)
= 2x2 − 3x2
= −x2
When x = 6, the value of the difference is
(f − g)(6) = −62 = −36.
6.1−6.4 What Did You Learn? (p. 325)
1. Calculating 10% of 233 million gives 23.3 million.
Because the number of cell phone users was increasing by
only 6%, the number of users should not increase by that
many in the several years after 2006.
2. Because the exponents in the equations in Exercises 23
and 26 are positive, the graphs must show growth. Because
the exponents in Exercises 24 and 25 are negative, the graphs
must show decay. Substituting 0 for x to fi nd the y-intercept
can also be used to match the graphs.
3. Find the coordinates when x = 1 and x = 10. Then fi nd
the rate of change. Compare the rates of each function.
6.1–6.4 Quiz (p. 326)
1. The function represents exponential growth because the
base, 4.25, is greater than 1.
2. The function represents exponential decay because the
base, 3 —
8 , is greater than 0 and less than 1.
3. The function represents exponential growth because the
base, e0.6, is greater than 1.
4. The function represents exponential decay because the
base, e−2, is greater than 0 and less than 1.
5. e8 ⋅ e4 = e8+4 = e12
6. 15e3 —
3e = 5e3−1 = 5e2
7. (5e4x)3 = 53e3(4x) = 125e12x
8. eln 9 = 9
9. log7 49x = log7(72)x = log7 7
2x = 2x
10. log3 81−2x = log3(34)−2x = log3 3
−8x = −8x
11. log4 1024 = 5, so 45 = 1024.
12. log1�3 27 = −3, so ( 1 — 3 )
−3
= 27.
13. 74 = 2401, so log7 2401 = 4.
14. 4−2 = 0.0625, so log4 0.0625 = −2.
15. log 45 ≈ 1.653
16. ln 1.4 ≈ 0.336
17. log2 32 = log2 25 = 5
Copyright © Big Ideas Learning, LLC Algebra 2 317All rights reserved. Worked-Out Solutions
Chapter 6
18. The inverse of f(x) is g(x) = log1�9 x.
y
6 8
6
8
x−2
−2
f(x) = (19)x
g(x) = log1/9 x
19. The inverse of y = ln(x − 7) is y = ex + 7.
y
1284 16
8
4
8
x−4
−4
16
812
g = ex + 7
y = ln (x − 7)
20. The inverse of f (x) = log5(x + 1) is g(x) = 5x − 1.
y
642
4
6
2
x−2
f(x) = log5 (x + 1)
g(x) = 5x − 1
21. The graph of g is a translation 2 units right and 1 unit up
of the graph of f. So, g(x) = log3(x − 2) + 1.
22. The graph of g is a translation 4 units down of the graph
of f. So, g(x) = 3x − 4.
23. The graph of g is a refl ection in the x-axis of the graph
of f. So, g(x) = −log1�2 x.
24. The growth rate of the value of the lamp is
1 + 0.0215 = 1.0215, so the model that represents the
value of the lamp is y = 150(1.0215)t.
25. a. For the 36-month CD at 2.0%, the value at the end of
the term is 1500 ( 1 + 0.02 —
12 )
12⋅3
≈ $1592.68, so about
$92.68 in interest was earned. For the 60-month
CD at 3.0%, the value at the end of the term is
2000 ( 1 + 0.03 —
12 ) 12⋅5
≈ $2323.23, so about $323.23 of
interest was earned.
b. The benefi t of the 36-month CD is the lesser minimum
balance, and the drawback is the lower interest rate. The
benefi t of the 60-month CD is the higher interest rate, and
the drawback is the greater minimum balance.
26.
20,000 40,0000 E
R
2
3
4
5
6
7
8
9
10
10
Energy (kilowatt-hours)
Ric
hte
r m
agn
itu
de
When E = 23,000, R = 0.67 ln 23,000 + 1.17 ≈ 7.9. So, the
Richter magnitude is about 7.9.
6.5 Explorations (p. 327)
1. logb mn = logb m + logb n
2. logb m
— n = logb m − logb n
3. logb mn = n logb m
4. Because expressions in exponential form can be rewritten in
logarithmic form, the properties of exponents can be used to
derive properties of logarithms.
5. a. log4 163 = 3 log4 16 = 3 ⋅ 2 = 6 b. log3 81−3 = −3 log3 81 = −3 ⋅ 4 = −12
c. ln e2 + ln e5 = ln (e2e5) = ln e7 = 7
d. 2 ln e6 − ln e5 = ln e12 − ln e5 = ln e12
— e5
= ln e7 = 7
e. log5 75 − log5 3 = log5 75
— 3 = log5 25 = 2
f. log4 2 + log4 32 = log4 (2 ⋅ 32) = log4 64 = 3
6.5 Monitoring Progress (pp. 328–330)
1. log6 5 —
8 = log6 5 − log6 8
≈ 0.898 − 1.161
= −0.263
2. log6 40 = log6 5 + log6 8
≈ 0.898 + 1.161
= 2.059
318 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
3. log6 64 = 2 log6 8
≈ 2 ⋅ 1.166
= 2.322
4. log6 125 = 3 log6 5
≈ 3 ⋅ 0.898
= 2.694
5. log6 3x 4 = log6 3 + log6 x4
= log6 3 + 4 log6 x
6. ln 5 —
12x = ln 5 − ln 12x
= ln 5 − (ln 12 + ln x)
= ln 5 − ln 12 − ln x
7. log x − log 9 = log x —
9
8. ln 4 + 3 ln 3 − ln 12 = ln 4 + ln 33 − ln 12
= ln 4 ⋅ 33 − ln 12
= ln 4 ⋅ 33
— 12
= ln 9
9. log5 8 = log 8 —
log 5
≈ 1.292
10. log8 14 = log 14 —
log 8
≈ 1.269
11. log26 9 = log 9 —
log 26
≈ 0.674
12. log12 30 = log 30 —
log 12
≈ 1.369
13. Let I be the original intensity, so that 3I is the tripled
intensity.
Increase in loudness = L(3I) − L(I)
= 10 log 3I
— I0
− 10 log I —
I0
= 10 ( log 3I
— I0
− log I —
I0
) = 10 ( log 3 + log
I —
I0
− log I —
I0
) = 10 log 3
The loudness increases by 10 log 3 decibels, or about
4.8 decibels.
6.5 Exercises (pp. 331–332)
Vocabulary and Core Concept Check
1. To condense the expression log3 2x + log3 y, use the Product
Property of Logarithms.
2. Use the Change-of-Base Formula with the common logarithm
or with the natural logarithm to evaluate the expression on
a calculator.
Monitoring Progress and Modeling with Mathematics
3. log7 3 = log7 12
— 4
= log7 12 − log7 4 ≈ 1.277 − 0.712
= 0.565
4. log7 48 = log7 (4 ⋅ 12)
= log7 4 + log7 12
≈ 0.712 + 1.277
= 1.989
5. log7 16 = log7 42 = 2 log7 4
≈ 2 ⋅ 0.712
= 1.424
6. log7 64 = log7 43 = 3 log7 4
≈ 3 ⋅ 0.712
= 2.136
7. log7 1 —
4 = log7 4
−1 = −log7 4 ≈ −(0.712)
= −0.712
8. log7 1 —
3 = log7
4 —
12
= log7 4 − log7 12
≈ 0.712 − 1.277
= −0.565
9. The expression matches B because
log3 6 − log3 2 = log3 6 —
2 = log3 3 by the Quotient Property
of Logarithms.
10. The expression matches D because
2 log3 6 = log3 62 = log3 36 by the Power Property
of Logarithms.
11. The expression matches A because
6 log3 2 = log3 26 = log3 64 by the Power Property
of Logarithms.
12. The expression matches C because
log 3 6 + log3 2 = log3 6 ⋅ 2 = log3 12 by the Product
Property of Logarithms.
Copyright © Big Ideas Learning, LLC Algebra 2 319All rights reserved. Worked-Out Solutions
Chapter 6
13. log3 4x = log3 4 + log3 x 14. log8 3x = log8 3 + log8 x
15. log 10x5 = log 10 + log x5 = 1 + 5 log x
16. ln 3x4 = ln 3 + ln x4 = ln 3 + 4 ln x
17. ln x —
3y = ln x − ln 3y
= ln x − ln 3 − ln y
18. ln 6x2
— y4
= ln 6x2 − ln y4 = ln 6 + ln x2 − ln y4 =
ln 6 + 2 ln x − 4 ln y
19. log7 5 √—
x = log7 5x1�2 = log7 5 + log7 x1�2 =
log7 5 + 1 — 2
log7 x
20. log5 3 √—
x2y = 1 — 3 log5 x2y = 1 —
3 log5 x2 + 1 —
3 log5 y =
2 — 3 log5 x + 1 —
3 log5 y
21. The error is the application of the Product Property, there
should be a sum rather than product.
log2 5x = log2 5 + log2 x
22. The error is in the location of the 3, it is on the wrong term.
ln 8x3 = ln 8 + 3 ln x
23. log4 7 − log4 10 = log4 7 —
10
24. ln 12 − ln 4 = ln 12
— 4 = ln 3
25. 6 ln x + 4 ln y = ln x6 + ln y4 = ln x6y4
26. 2 log x + log 11 = log x2 + log 11 = log 11x2
27. log5 4 + 1 — 3
log5 x = log5 4 + log5 3 √—
x = log5 4 3 √—
x
28. 6 ln 2 − 4 ln y = ln 26 − ln y4 = ln 64 − ln y4 = ln 64
— y4
29. 5 ln 2 + 7 ln x + 4 ln y = ln 25 + ln x7 + ln y4 = ln 32 + ln x7 + ln y4 = ln 32 x7y4
30. log3 4 + 2 log3 1 —
2 + log3 x = log3 4 + log3
1 —
4 + log3 x
= log3 4x
— 4
= log3 x
31. B;
log5 y4
— 3x
= log5 y4 − log5 3x Quotient Property
= 4 log5 y − (log5 3 + log5 x) Power and Product
Properties
= 4 log5 y − log5 3 − log5 x Distributive
Property
32. B;
9 log x − 2 log y = log x9 − log y2 Power Property
= log x9
— y2
Quotient Property
33. log4 7 = log 7 —
log 4 ≈ 1.404 34. log5 13 = log 13 —
log 5 ≈ 1.594
35. log9 15 = log 15 —
log 9 ≈ 1.232 36. log8 22 = log 22 —
log 8 ≈ 1.486
37. log6 17 = log 17 —
log 6 ≈ 1.581 38. log2 28 = log 28 —
log 2 ≈ 4.807
39. log7 3 —
16 =
log 3 — 16 —
log 7 ≈ −0.860
40. log3 9 —
40 =
log 9 — 40 —
log 3 ≈ −1.358
41. Your friend is correct. By the change-of-base formula, use
y = log x —
log 3 in place of y = log3 x.
42. The value of log 8
— log 2
is 3.
43. Let I be the intensity of the loudest sound that a human can
make, so 1,000,000I is the intensity of the sound a blue
whale can produce.
Difference in loudness = L(1,000,000I) − L(I)
= 10 log 1,000,000I —
I0
− 10 log I —
I0
= 10 ( log 1,000,000I —
I0
− log I — I0
) = 10 ( log 1,000,000 + log I —
I0
− log I — I0
) = 10 ⋅ 6
= 60
The difference in decibel levels is 60 decibels.
44. Let I be the intensity of the sound of the show, so the
intensity of the sound of the advertisement is 10I.
Increase in loudness = L(10I) − L(I)
= 10 log 10I —
I0
− 10 log I — I0
= 10 ( log 10I —
I0
− log I — I0
) = 10 ( log 10 + log I —
I0
− log I — I 0
) = 10(1)
= 10
The loudness increases by 10 decibels.
320 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
45. a. Let h be the original height, so that 2h is the doubled
height.
Increase in speed = s(2h) − s(h)
= 2 ln 2 ⋅ 100h − 2 ln 100h
= 2(ln 2 ⋅ 100h − ln 100h)
= 2(ln 2 + ln 100h − ln 100h)
= 2 ln 2
The speed increases by 2 ln 2 knots, or about 1.39 knots.
b. s(h) = 2 ln 100h
= 2 ( log 100h —
log e )
= 2 —
log e (log 100h)
= 2 —
log e (log h + log 100)
= 2 —
log e (log h + 2)
46. The formula is not true. By the Product Property,
logb M + logb N = logb (MN). If logb (M + N) = logb (MN),
then M + N = M ⋅ N which is not always true.
47. Because x = logb a, y = logb c, and z = logc a, it follows
that bx = a, by = c, and cz = a. By substitution, (by)z = a,
and then (by)z = bx. So, byz = bx and yz = x. So, z = x — y and
logc a = logb a — logb c
.
48. Sample answer: The graph of g could be a translation 1 unit
up of the graph of f. The graph of g could be a horizontal
shrink by a factor of 1 —
100 followed by a translation 1 unit
down of the graph of f. The graph of g could be a horizontal
shrink by a factor of 1 —
10 of the graph of f;
Transformation 1:
g(x) = f (x) + 1 = log x + 1 = 2 + log x − 1 = log 100 + log x − 1 = log 100x − 1;
Transformation 2: g(x) = f (100x) − 1 = log 100x − 1;
Transformation 3: g(x) = f (10x) = log(10x) = log ( 100x —
10 )
= log 100x − log 10 = log 100x − 1
Maintaining Mathematical Profi ciency
49. The solution consists of the x-values for which the graph of y = x2 − 4 lies above the x-axis. Find the x-intercepts of the
graph by letting y = 0 and solving for x.
y = x2 − 4 0 = x2 − 4 4 = x2
±2 = x The solutions are x = −2 and x = 2. Sketch a parabola that
opens up and has −2 and 2 as x-intercepts.
x
y
2
4
−2
−4 4
The graph lies above the x-axis to the left of x = −2 and to
the right of x = 2. The solution of the inequality is x < −2
or x > 2.
50. Rewrite the inequality.
2(x − 6)2 − 5 ≥ 37
2(x − 6)2 − 42 ≥ 0 The solution consists of the x-values for which the graph
of y = 2(x − 6)2 − 42 lies on or above the x-axis. Find the
x-intercepts of the graph by letting y = 0 and solving for x.
y = 2(x − 6)2 − 42
0 = 2(x − 6)2 − 42
42 = 2(x − 6)2
21 = (x − 6)2
± √—
21 = x − 6 6 ± √
— 21 = x
The solutions are x ≈ −1.42 and x ≈ 10.58. Sketch
a parabola that opens up and has −1.42 and 10.58 as
x-intercepts.
x
y
20
−20
−40
4 8 12
The graph lies on or above the x-axis to the left of x ≈ −1.42
and to the right of x ≈ 10.58. The solution of the inequality
is x ≤ 6 − √—
21 or x ≥ 6 + √—
21 .
Copyright © Big Ideas Learning, LLC Algebra 2 321All rights reserved. Worked-Out Solutions
Chapter 6
51. The solution consists of the x-values for which the graph
of y = x2 + 13x + 42 lies below the x-axis. Find the
x-intercepts of the graph by letting y = 0 and solving for x.
y = x2 + 13x + 42
0 = x2 + 13x + 42
0 = (x + 7)(x + 6)
x + 7 = 0 or x + 6 = 0 x = −7 x = −6
Sketch a parabola that opens up and has −7 and −6 as
x-intercepts.
x
y
2
4
6
−4−6−8 −2
The graph lies below the x-axis between x = −7 and x = −6. The solution of the inequality is −7 < x < −6.
52. Rewrite the inequality.
−x2 − 4x + 6 ≤ −6
−x2 − 4x + 12 ≤ 0 The solution consists of the x-values for which the graph
of y = −x2 − 4x + 12 lies on or below the x-axis. Find the
x-intercepts of the graph by letting y = 0 and solving for x.
y = −x2 − 4x + 12
0 = −x2 − 4x + 12
0 = x2 + 4x − 12
0 = (x + 6)(x − 2)
x + 6 = 0 or x − 2 = 0 x = −6 x = 2 Sketch a parabola that opens down and has −6 and 2
as intercepts.
x
y
4
8
−8 −4 84
12
The graph lies on or below the x-axis to the left of x = −6
and to the right of x = 2. The solution of the inequality is x ≤ −6 or x ≥ 2.
53. Step 1 Write a system of equations using each side of the
original equation.
Equation System
4x2 − 3x − 6 = −x2 + 5x + 3 y = 4x2 − 3x − 6 y = −x2 + 5x + 3 Step 2 Use a graphing calculator to graph the system. Then
use the intersect feature to fi nd the x-value of each
solution of the system.
10
−8
−10
10
The graphs intersect when x ≈ 2.36 and x ≈ −0.76.
The solutions of the equation are x ≈ 2.36 and x ≈ −0.76.
54. Step 1 Write a system of equations using each side of the
original equation.
Equation System
−(x + 3)(x − 2) = x2 − 6x y = −(x + 3)(x − 2)
y = x2 − 6x
Step 2 Use a graphing calculator to graph the system. Then
use the intersect feature to fi nd the x-value of each
solution of the system.
8
−10
−6
8
The graphs intersect when x ≈ 3.39 and x ≈ −0.89.
The solutions of the equation are x ≈ 3.39 and x ≈ −0.89.
55. Step 1 Write a system of equations using each side of the
original equation.
Equation System
2x2 − 4x − 5 = −(x + 3)2 + 10 y = 2x2 − 4x − 5 y = −(x + 3)2 + 10
Step 2 Use a graphing calculator to graph the system. Then
use the intersect feature to fi nd the x-values of each
solution of the system.
10
−8
−10
12
The graphs intersect when x ≈ 1.12 and x ≈ −1.79.
The solutions of the equation are x ≈ 1.12 and x ≈ −1.79.
322 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
56. Step 1 Write a system of equations using each side of the
original equation.
Equation System
−(x + 7)2 + 5 = (x + 10)2 − 3 y = −(x + 7)2 + 5 y = (x + 10)2 − 3 Step 2 Use a graphing calculator to graph the system. Then
use the intersect feature to fi nd the x-value of each
solution of the system.
2
−4
−13
6
The graphs intersect when x ≈ −7.177 and x ≈ −9.823.
The solutions of the equation are x ≈ −7.177 and x ≈ −9.823.
6.6 Explorations (p. 333)
1. a. C; The graph shows an exponential growth function with a
constant function at y = 2; x ≈ 0.7
b. F; The graph shows a logarithmic function with a constant
function at y = −1; x ≈ 0.4
c. A; The graph shows an exponential growth and an
exponential decay function; x = 0 d. E; The graph shows a logarithmic function with a constant
function at y = 1; x = 4
e. B; The graph shows a logarithmic function with a constant
function at y = 1 — 2 ; x = √—
5 ≈ 2.24
f. D; The graph shows an exponential growth function with
a constant function at y = 2; x = 1 — 2
2. a. Sample answer: For ex = 2, use ex = y:
x 0.55 0.60 0.65 0.70 0.75 0.80
y 1.7333 1.8221 1.9155 2.0138 2.1170 2.2255
Because e0.70 ≈ 2, x ≈ 0.70; For ln x = −1, use ln x = y:
x 0.32 0.34 0.36 0.38 0.40 0.42
y −1.139 −1.079 −1.022 −0.968 −0.916 −0.868
Because ln 0.36 ≈ −1, x ≈ 0.36.
b. Sample answer:
ex = 2 ln ex = ln 2
x ln e = ln 2
x = ln 2
x ≈ 0.69;
ln x = −1
eln x = e−1
x = e−1
x ≈ 0.37
3. Approximate answers can be found using a graph or table
of values. Exact answers can be found by using the inverse
properties of logarithms and exponents.
4. Sample answer: Use inverse properties to fi nd exact
solutions.
a. 16x = 2 log16 16x = log16 2
x = log16 2
x = 1 — 4
The solution is x = 1 — 4 .
b. 2x = 42x+1
2x = 22(2x+1)
x = 2(2x + 1)
x = 4x + 2 −3x = 2 x = − 2 —
3
The solution is x = − 2 — 3 .
c. 2x = 3 x+1
x ln 2 = (x + 1) ln 3
x ln 2 = x ln 3 + ln 3
x ln 2 − x ln 3 = ln 3
x (ln 2 − ln 3) = ln 3
x ln 2 —
3 = ln 3
x = ln 3 ____
ln 2 —
3
The solution is x = ln 3 ____
ln 2 —
3 ≈ −2.71.
d. log x = 1 — 2
10log x = 101�2
x = 101�2
The solution is x = 101�2 = √—
10 ≈ 3.16.
e. ln x = 2 eln x = e2
x = e2
The solution is x = e2 ≈ 7.39.
f. log3 x = 3 —
2
3log3 x = 33�2
x = 33�2
The solution is x = 33�2 ≈ 5.20.
Copyright © Big Ideas Learning, LLC Algebra 2 323All rights reserved. Worked-Out Solutions
Chapter 6
6.6 Monitoring Progress (pp. 334–337)
1. 2x = 5 2x = 2log2 5
x = log2 5 x ≈ 2.322
2. 79x = 15
9x = log7 15
x = 1 — 9 log7 15
x ≈ 0.155
3. 4e−0.3x − 7 = 13
4e−0.3x = 20
e−0.3x = 5 −0.3x = ln 5
x = − 1 —
0.3 ln 5
x ≈ −5.365
4. Use Newton’s Law of Cooling with T = 100, T0 = 212,
TR = 75, and r = 0.046.
T = (T0 − TR) e−rt + TR
100 = (212 − 75)e−0.046t + 75
25 = 137e−0.046t
0.182 ≈ e−0.046t
ln 0.182 ≈ ln e−0.046t
−1.701 ≈ −0.046t
36.9 ≈ t You should wait about 37 minutes before serving the stew.
5. ln (7x − 4) = ln (2x + 11) Check
7x − 4 = 2x + 11 ln (7 ⋅ 3 − 4) =?
ln (2 ⋅ 3 + 11)
7x = 2x + 15 ln 17 = ln 17 ✓
5x = 15
x = 3
6. log2 (x − 6) = 5 Check
2 log 2 (x−6) = 25 log2 (38 − 6) =
? 5
x − 6 = 32 log2 32 =?
5 x = 38 5 = 5 ✓
7. log 5x + log(x − 1) = 2
log [5x(x − 1)] = 2
10log[5x(x−1)] = 102
5x(x − 1) = 100
5x2 − 5x = 100
5x2 − 5x − 100 = 0
x2 − x − 20 = 0
(x − 5)(x + 4) = 0
x − 5 = 0 or x + 4 = 0
x = 5 x = −4
Check
log 5 ⋅ 5 + log(5 − 1) =?
2
log 25 + log 4 =?
2
log 100 =?
2
2 = 2 ✓
log 5(−4) + log(−4 − 1) =?
2
log (−20) + log −5 =?
2
Because log (−20) is not defi ned, −4 is not a solution. ✗
The apparent solution x = −4 is extraneous. So, the only
solution is x = 5.
8. log4(x + 12) + log4x = 3
log4[x(x + 12)] = 3
4log4[x(x+12)] = 43
x(x + 12) = 64
x2 + 12x = 64
x2 + 12x − 64 = 0
(x − 4)(x + 16) = 0
x − 4 = 0 or x + 16 = 0
x = 4 x = −16
Check
log4(4 + 12) + log44 =?
3
log416 + log44 =?
3
2 + 1 =?
3
3 = 3 ✓
log4(−16 + 12) + log4 −16 =?
2
Because log4 (−16) is not defi ned, −16 is not a solution. ✗
The apparent solution x = −16 is extraneous. So, the only
solution is x = 4.
9. ex < 2
ln ex < ln 2
x < ln 2
The solution is x < ln 2. Because ln 2 ≈ 0.693, the
approximate solution is x < 0.693.
324 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
10. 102x−6 > 3
log(102x−6) > log 3
2x − 6 > log 3
2x > log 3 + 6
x > 1 — 2 log 3 + 3
The solution is x > 1 —
2 log 3 + 3. Because
1 —
2 log 3 + 3 ≈
3.239, the approximate solution is x > 3.239.
11. log x + 9 < 45
log x < 36
10log x < 1036
x < 1036
Because log x is only defi ned when x > 0, the solution is
0 < x < 1036.
12. 2 ln x − 1 > 4
2 ln x > 5
ln x > 5 — 2
eln x > e5/2
x > e5/2
The solution is x > e5/2. Because e5/2 ≈ 12.182, the
approximate solution is x > 12.182.
6.6 Exercises (pp. 338 –340)
Vocabulary and Core Concept Check
1. The equation 3 x−1 = 34 is an example of an exponential
equation.
2. When solving exponential equations, the exponents can be
set equal once a common base is found. If the bases are not
the same, try solving the equation by taking a logarithm of
each side. When solving logarithmic functions, each side of
the equation can be exponentiated to obtain an equation with
no logarithms.
3. The domain of a logarithmic function is positive numbers
only, so any quantity that results in taking the log of a non-
positive number will be an extraneous solution.
4. If b is a positive real number other than l, then bx = by if and
only if x = y.
Monitoring Progress and Modeling with Mathematics
5. 73x+5 = 7 1−x
3x + 5 = 1 − x
4x + 5 = 1
4x = −4
x = −1
6. e2x = e3x−1
2x = 3x − 1
−x = −1
x = 1
7. 5x−3 = 25x−5
5x−3 = 52(x−5)
x − 3 = 2(x − 5)
x − 3 = 2x − 10
−x − 3 = −10
−x = −7
x = 7
8. 62x−6 = 363x−5
62x−6 = 62(3x−5)
2x − 6 = 2(3x − 5)
2x − 6 = 6x − 10
−4x − 6 = −10
−4x = −4
x = 1
9. 3x = 7
log3 3x = log3 7
x = log3 7 x ≈ 1.771
10. 5x = 33
log55x = log5 33
x = log5 33
x ≈ 2.173
11. 495x+2 = ( 1 — 7 )
11−x
72(5x+2) = 7−1(11−x)
2(5x + 2) = −1(11 − x)
10x + 4 = −11 + x
9x + 4 = −11
9x = −15
x = − 5 — 3
12. 5125x−1 = ( 1 — 8 )
−4−x
29(5x−1) = 2−3(−4−x)
9(5x − 1) = −3(−4 − x)
45x − 9 = 12 + 3x
42x − 9 = 12
42x = 21
x = 1 — 2
13. 75x = 12
log7(75x) = log7 12
5x = log7 12
x = 1 — 5 log7 12
x ≈ 0.255
Copyright © Big Ideas Learning, LLC Algebra 2 325All rights reserved. Worked-Out Solutions
Chapter 6
14. 116x = 38
log11116x = log11 38
6x = log11 38
x = 1 — 6 log11 38
x ≈ 0.253
15. 3e4x + 9 = 15
3e4x = 6
e4x = 2
ln e4x = ln 2
4x = ln 2
x = 1 — 4 ln 2
x ≈ 0.173
16. 2e2x − 7 = 5
2e2x = 12
e2x = 6
ln e2x = ln 6
2x = ln 6
x = 1 — 2 ln 6
x ≈ 0.896
17. l = 266 − 219 e−0.05t
175 = 266 − 219e−0.05t
−91 = −219e−0.05t
0.416 ≈ e−0.05t
ln 0.416 ≈ ln e−0.05t
−0.878 ≈ −0.05t
17.56 ≈ t
The shark is about 17.6 years old.
18. R = 100e−0.00043t
5 = 100e−0.00043t
0.05 = e−0.00043t
ln 0.05 = ln e−0.00043t
−2.996 ≈ −0.00043t
6966.82 ≈ t
It will take about 6967 years for these to be only 5 grams of
radium.
19. Use Newton’s Law of Cooling and let T = 230, T0 = 280,
TR = 80, and r = 0.0058.
T = (T0 − TR)e−rt + TR
230 = (280 − 80)e−0.0058t + 80
150 = 200e−0.0058t
0.75 = e−0.0058t
ln 0.75 = ln e−0.0058t
−0.288 ≈ −0.0058t
49.6 ≈ t
You have to wait about 50 minutes before you can continue
driving.
20. Use Newton’s Law of Cooling and let T = 100, T0 = 180,
TR = 72, and r = 0.067.
T = (T0 − TR)e−rt + TR
100 = (180 − 72)e−0.067t + 72
28 = 108e−0.067t
0.259 ≈ e−0.067t
ln 0.259 ≈ ln e−0.067t
−1.35 ≈ −0.067t
20.1 ≈ t
You should wait about 20 minutes before carving the turkey.
21. ln(4x − 7) = ln(x + 11) Check
4x − 7 = x + 11 ln(4 ⋅ 6 − 7) =?
ln(6 + 11)
3x − 7 = 11 ln 17 = ln 17 ✓
3x = 18
x = 6
22. ln(2x − 4) = ln(x + 6) Check
2x − 4 = x + 6 ln(2 ⋅ 10 − 4) =?
ln(10 + 6)
x − 4 = 6 ln(16) = ln(16) ✓
x = 10
23. log2(3x − 4) = log25 Check
3x − 4 = 5 log2(3 ⋅ 3 − 4) =?
log2 5
3x = 9 log2 5 =?
log2 5 ✓
x = 3
24. log(7x + 3) = log 38 Check
7x + 3 = 38 log(7 ⋅ 5 + 3) =?
log 38
7x = 35 log 38 = log 38 ✓
x = 5
25. log2(4x + 8) = 5 Check
2 log2(4x+8) = 25 log2(4 ⋅ 6 + 8) =?
5 4x + 8 = 32 log2(32) =
? 5
4x = 24 5 =?
5 ✓
x = 6
326 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
26. log3(2x + 1) = 2 Check
3 log3(2x+1) = 32 log3(2 ⋅ 4 + 1) =?
2 2x + 1 = 9 log3 9 =
? 2
2x = 8 2 =?
2 ✓
x = 4
27. log7(4x + 9) = 2 Check
7 log7(4x+9) = 72 log7(4 ⋅ 10 + 9) =?
2 4x + 9 = 49 log7 49 =
? 2
4x = 40 2 = 2 ✓
x = 10
28. log5(5x + 10) = 4 Check
5 log5(5x+10) = 54 log5(5 ⋅ 123 + 10) =?
4 5x + 10 = 625 log5 625 =
? 4
5x = 615 4 = 4 ✓
x = 123
29. log(12x − 9) = log3x Check
10log(12x−9) = 10log3x log(12 ⋅ 1 − 9) =?
log 3 ⋅ 1
12x − 9 = 3x log 3 = log 3 ✓
9x − 9 = 0 9x = 9 x = 1
30. log6(5x + 9) = log66x Check
6 log6(5x+9) = 6 log66x log6 (5 ⋅ 9 + 9) = log66 ⋅ 9
5x + 9 = 6x log6 54 = log6 54 ✓
−x + 9= 0 −x = −9
x = 9
31. log2(x2 − x − 6) = 2 2 lo g2(x 2 −x−6) = 22
x2 − x − 6 = 4 x2 − x − 10 = 0
x = 1 ± √——
1 − 4(1)(−10) ——
2
x = 1 ± √—
41 —
2
x ≈ 3.7 or x ≈ −2.7
Check
log2(3.72 − 3.7 − 6) =?
2 log2(3.99) =
? 2
1.996 ≈ 2 ✓
log2 [ (−2.7)2 − (−2.7) − 6 ] =?
2
log2(3.99) =?
2 1.996 ≈ 2 ✓
32. log3(x2 + 9x + 27) = 2 3 lo g3(x 2 +9x+27) = 32
x2 + 9x + 27 = 9 x2 + 9x + 18 = 0 (x + 3)(x + 6) = 0 x + 3 = 0 or x + 6 = 0 x = −3 x = −6
Check
log3 [ (−3)2 + 9(−3) + 27 ] =?
2
log3 9 =?
2 2 = 2 ✓
log3 [ (−6)2 + 9(−6) + 27 ] =?
2
log3 9 =?
2
2 = 2 ✓
33. log2 x + log2 (x − 2) = 3 log2[x(x − 2)] = 3 2 log
2 [x(x−2)] = 23
x(x − 2) = 8 x2 − 2x = 8 x2 − 2x − 8 = 0 (x − 4)(x + 2) = 0 x − 4 = 0 or x + 2 = 0 x = 4 x = −2
Check
log2 4 + log2(4 − 2) =?
3
log2 4 + log2 2 =?
3
2 + 1 =?
3 3 = 3 ✓
log2(−2) + log2(−2 − 2) =?
3 Because log2 (−2) is not defi ned, −2 is not a solution.
The apparent solution x = −2 is extraneous. So, the only
solution is x = 4.
Copyright © Big Ideas Learning, LLC Algebra 2 327All rights reserved. Worked-Out Solutions
Chapter 6
34. log6 3x + log6(x − 1) = 3 log6[3x(x − 1)] = 3 6 log6[3x(x−1)] = 63
3x(x − 1) = 216
3x2 − 3x = 216
x2 − x = 72
x2 − x − 72 = 0 (x − 9)(x + 8) = 0 x − 9 = 0 or x + 8 = 0 x = 9 x = −8
Check
log6 3 ⋅ 9 + log6(9 − 1) =?
3 log6 27 + log6 8 =
? 3
log6 216 =?
3 3 = 3 ✓
log6 3(−8) + log6(−8 − 1) =?
3 log6(−24) + log6(−9) =
? 3
Because log6(−24) is not defi ned, −8 is not a solution.
The apparent solution x = −8 is extraneous. So, the only
solution is x = 9.
35. ln x + ln(x + 3) = 4 ln[x(x + 3)] = 4 eln[x(x+3)] = e4
x(x + 3) = e4
x2 + 3x = e4
x2 + 3x − e4 = 0 x ≈ 6.039 or x ≈ −9.039
Check
ln(−9.039) + ln(−9.039 + 3) =?
4 Because ln(−9.039) is not defi ned, −9.039 is not a solution.
ln 6.039 + ln(6.039 + 3) =?
4 1.7982 + 2.2015 ≈ 4
3.9997 ≈ 4 ✓
The approximate apparent solution x ≈ −9.039 is
extraneous. So, the only solution is x ≈ 6.039.
36. ln x + ln(x − 2) = 5
ln[x(x − 2)] = 5
eln[x(x−2)] = e5
x(x − 2) = e5
x2 − 2x = e5
x2 − 2x − e5 = 0
x ≈ 13.223 or x ≈ −11.223
Check
ln(−11.223) + ln(−11.223 − 2) =?
5
Because ln(−11.223) is not defi ned, −11.223 is not
a solution.
ln 13.223 + ln(13.223 − 2) =?
5
ln 13.223 + ln(11.223) =?
5
2.582 + 2.418 =?
5
5 = 5 ✓
The apparent solution x ≈ −11.223 is extraneous. So, the
only solution is x ≈ 13.223.
37. log3 3x2 + log3 3 = 2
log3 9x2 = 2
3log 3 9x 2 = 32
9x2 = 9
x2 = 1
x = ±1
Check
log3 3(−1)2 + log3 3 =?
2
log3 3 + log3 3 =?
2
1 + 1 =?
2 2 = 2 ✓
log3 3 ⋅ 12 + log3 3 =?
2
log3 3 + log3 3 =?
2
1 + 1 =?
2 2 = 2 ✓
The solutions are x = −1 and x = 1.
328 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
38. log4 (−x) + log4 (x + 10) = 2 log4[−x(x + 10)] = 2 4 lo g
4 [−x(x+10)] = 42
−x(x + 10) = 16
−x2 − 10x = 16
−x2 − 10x − 16 = 0 x2 + 10x + 16 = 0 (x + 2)(x + 8) = 0 x + 2 = 0 or x + 8 = 0 x = −2 x = −8
Check
log4 [ −(−2) ] + log4 (−2 + 10) =?
2 log4 2 + log4 8 =
? 2
log4 16 =?
2 2 = 2 ✓
log4 [ −(−8) ] + log4(−8 + 10) =?
2
log4 8 + log4 2 =?
2 log4 16 =
? 2
2 = 2 ✓
The solutions are x = −2 and x = −8.
39. log3 (x − 9) + log3(x − 3) = 2 log3[(x − 9)(x − 3)] = 2 3lo g
3 [(x−9) (x−3)] = 32
(x − 9)(x − 3) = 9 x2 − 12x + 27 = 9 x2 − 12x + 18 = 0 x ≈ 10.24 or x ≈ 1.76
Check
log3 (1.76 − 9) + log3 (1.76 − 3) =?
2 log3 (−7.24) + log3 (−1.24) =
? 2
Because log3 (−7.24) is not defi ned, 1.76 is not a solution.
log3 (10.24 − 9) + log3 (10.24 − 3) =?
2
log3 1.24 + log3 7.24 =?
2 0.1958 + 1.8019 ≈ 2 ✓
The apparent solution x ≈ 1.76 is extraneous. So, the only
solution is x ≈ 10.24.
40. log5 (x + 4) + log5 (x + 1) = 2 log5 [(x + 4)(x + 1)] = 2 5 log5[(x+4)(x+1)] = 52
(x + 4)(x + 1) = 25
x2 + 5x + 4 = 25
x2 + 5x − 21 = 0 x ≈ 2.72 or x ≈ −7.72
Check
log5(−7.72 + 4) + log5(−7.72 + 1) =?
2 log5(−3.72) + log5(−6.72) =
? 2
Because log5(−3.72) is not defi ned, −7.72 is not a solution.
log5(2.72 + 4) + log5(2.72 + 1) =?
2 log5(6.72) + log5(3.72) =
? 2
1.184 + 0.816 =?
2
2 = 2 ✓
The apparent solution x ≈ −7.72 is extraneous. So, the only
solution is x ≈ 2.72.
41. The error is that the base of 3 was not used on both sides of
the equation.
log3(5x − 1) = 4 3 log3(5x−1) = 34
5x − 1 = 81
5x = 80
x = 20
42. The error is that the apparent solutions were not
checked.
Check
log4(−16 + 12) + log4(−16) =?
3
Because log4 (−16) is not defi ned, −16 is not a solution.
log4(4 + 12) + log4 4 =?
3
log4 16 + log4 4 =?
3
2 + 1 =?
3
3 = 3 ✓
The only solution is x = 4.
43. a. A = P ( 1 + r — n )
nt
1000 = 100(1 + 0.06)t
10 = (1.06)t
ln 10 = ln(1.06)t
ln 10 = t ln 1.06
ln 10
— ln 1.06
= t
39.52 ≈ t It will take about 39.52 years.
Copyright © Big Ideas Learning, LLC Algebra 2 329All rights reserved. Worked-Out Solutions
Chapter 6
b. A = P ( 1 + r — n )
nt
1000 = 100 ( 1 + 0.06 —
4 )
4t
10 = (1.015)4t
ln 10 = ln(1.015)4t
ln 10 = 4t ln(1.015)
ln 10 —
4 ln (1.015) = t
38.66 ≈ t It will take about 38.66 years.
c. A = P ( 1 + r — n )
nt
1000 = 100 ( 1 + 0.06 —
365 )
365t
10 ≈ (1.0002)365t
ln 10 ≈ ln(1.0002)365t
ln 10 ≈ 365t ln(1.0002)
ln 10 ——
365 ln(1.0002) ≈ t
38.38 ≈ t It will take about 38.38 years.
d. A = Pert
1000 = 100e0.06t
10 = e0.06t
ln 10 = ln e0.06t
ln 10 = 0.06t
ln 10 —
0.06 = t
38.38 ≈ t It will take about 38.38 years.
44. M = 5 log D + 2 12 = 5 log D + 2 10 = 5 log D
2 = log D
102 = 10logD
100 = D The diameter is 100 millimeters.
45. a. The solution is approximately x ≈ 3.8.
b. The solution is x = 0.8.
46. Your friend is incorrect. For example log4(−x) = 4 has
the solution x = −1.
47. 9x > 54
log9 9x > log9 54
x log9 9 > log9 54
x > log9 54
The solution is x > log9 54. Because log9 54 ≈ 1.815, the
approximate solution is x > 1.815.
48. 4x ≤ 36
log4 4x ≤ log4 36
x ≤ log4 36
The solution is x ≤ log4 36. Because log4 36 ≈ 2.585, the
approximate solution is x ≤ 2.585.
49. ln x ≥ 3
eln x ≥ e3
x ≥ e3
The solution is x ≥ e3. Because e3 ≈ 20.086, the
approximate solution is x ≥ 20.086.
50. log4 x < 4 4 log
4 x < 44
x < 256
Because log4 x is only defi ned when x > 0, the solution is
0 < x < 256.
51. 34x−5 < 8 log3(3
4x−5) < log3 8 4x − 5 < log3 8 4x < log3 8 + 5
x < 1 — 4 (log3 8 + 5)
The solution is x < 1 — 4 (log3 8 + 5). Because
1 —
4 (log3 8 + 5)
≈ 1.723, the approximate solution is x < 1.723.
52. e3x+4 > 11
ln e3x+4 > ln 11
3x + 4 > ln 11
3x > ln 11 − 4
x > 1 — 3 (ln 11 − 4)
The solution is x > 1 — 3 (ln 11 − 4). Because
1 —
3 (ln 11 − 4)
≈ −0.534, the approximate solution is x > −0.534.
53. −3 log5 x + 6 ≤ 9 −3 log5 x ≤ 3 log5 x ≥ −1
5 log5 x ≥ 5−1
x ≥ 1 —
5
The solution is x ≥ 1 —
5 .
330 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
54. −4 log5 x − 5 ≥ 3
−4 log5 x ≥ 8
log5 x ≤−2
5 log5 x ≤ 5−2
x ≤ 1 — 25
Because log5 x is only defi ned when x > 0, the solution is
0 < x ≤ 1 — 25
.
55. log5 x < 2 5 log5 x < 52
x < 25
x
y
1
3
−1
−3
12 18 246
y = log5 x
y = 2
Sample answer: The algebraic method is preferred since the
result is more accurate.
56. 1200 ≤ 1000 ( 1 + 0.035 —
12 )
12t
1.2 ≤ (1.0029)12t
ln 1.2 ≤ ln(1.0029)12t
ln 1.2 ≤ 12t ln(1.0029)
ln 1.2 ——
12 ln(1.0029) ≤ t
5.22 ≤ t
It will take about 5.22 years to reach $1200.
3500 ≤ 1000 ( 1 + 0.035 —
12 )
12t
3.5 ≤ (1.00292)12t
ln 3.5 ≤ ln(1.00292)12t
ln 3.5 ≤ 12t ln(1.00292)
ln 3.5 ——
12 ln(1.00292) ≤ t
35.85 ≤ t It will take about 35.85 years to reach $3500.
57. 10 = ln 2 —
ln(1 + r)
10 ln(1 + r) = ln 2
e10 ln(1+r) = eln2
(1 + r)10 = 2
1 + r = 2 1/10
r = 2 1/10 − 1 r ≈ 0.0718
The rate of return is about 0.0718 = 7.18%.
58. 10,000 ≤ 20,000 (1 − 0.15)t
1 —
2 ≤ (0.85)t
ln 1 —
2 ≤ ln(0.85)t
ln 1 —
2 ≤ t ln(0.85)
ln
1 —
2 _______
ln(0.85) ≤ t
4.27 ≤ t The value is greater than $10,000 for about 4.27 years.
59.
60
0
4
Using the intersect feature, the solution is x ≈ 1.782.
60.
5
−2
−1
2
Using the intersect feature, the solution is x ≈ 1.233.
61.
3
−1
0
1
Using the intersect feature, there is no solution.
62.
6
−1
0
3
Using the intersect feature, the solutions are x ≈ 3.693 and
x ≈ 0.545.
63. a. ℓ = 45 − 25.7e−0.09a
ℓ − 45 = −25.7e−0.09a
45 − ℓ —
25.7 = e−0.09a
ln ( 45 −ℓ — 25.7
) = −0.09a
1 —
−0.09a ln ( 45 −ℓ —
25.7 ) = a
Copyright © Big Ideas Learning, LLC Algebra 2 331All rights reserved. Worked-Out Solutions
Chapter 6
b. ℓ = 24: a = 1 — −0.01
ln ( 45 − 24 —
25.7 )
≈ 2.2 years
ℓ = 28: a = 1 — −0.09
ln ( 45 − 28 —
25.7 )
≈ 4.6 years
ℓ = 32: a = 1 — −0.09
ln ( 45 − 32 —
25.7 )
≈ 7.6 years
ℓ = 36: a = 1 — −0.09
ln ( 45 − 36 —
25.7 )
≈ 11.7 years
64. The solution is x > 2.1 because the graph of y = 4 ln x + 6
is above the graph of y = 9 to the right of x ≈ 2.1.
65. Sample answer: Exponential equation: 2x = 16
Logarithmic equation: log3 (−x) = 1
66. Sample answer: One solution: ln (x + 1) = 5 Two solutions: e x
2+1 = 10
No solution: ex = −1
67. 2x+3 = 53x−1
ln (2x+3) = ln (53x−1)
(x + 3) ln 2 = (3x − 1) ln 5
x ln 2 + 3 ln 2 = 3x ln 5 − ln 5
x ln 2 − 3x ln 5 = −3 ln 2 − ln 5
x (ln 2 − 3 ln 5) = −3 ln 2 − ln 5
x = −3 ln 2 − ln 5 ——
ln 2 − 3 ln 5
The solution is x = −3 ln 2 − ln 5 ——
ln 2 − 3 ln 5 . Because
−3 ln 2 − ln 5 ——
ln 2 − 3 ln 5 ≈
0.892, the approximate solution is x ≈ 0.892.
68. 103x−8 = 25−x
ln 103x−8 = ln 25−x
(3x − 8) ln 10 = (5 − x) ln 2
3x ln 10 − 8 ln 10 = 5 ln 2 − x ln 2
3x ln 10 + x ln 2 = 5 ln 2 + 8 ln 10
x(3 ln 10 + ln 2) = 5 ln 2 + 8 ln 10
x = 5 ln 2 + 8 ln 10 ——
3 ln 10 + ln 2
The solution is x = 5 ln 2 + 8 ln 10 ——
3 ln 10 + ln 2 . Because
5 ln 2 + 8 ln 10
—— 3 ln 10 + ln 2
≈ 2.879, the approximate solution
is x ≈ 2.879.
69. log3(x − 6) = log9 2x
9lo g 3 (x−6) = 9lo g
9 2x
[ 3lo g 3 (x−6) ] 2 = 2x
(x − 6)2 = 2x
x2 − 12x + 36 = 2x
x2 − 14x + 36 = 0 x ≈ 10.61 or x ≈ 3.39
Because log3 (−2.6) is not defi ned, the apparent solution
x ≈ 3.39 is an extraneous solution. The only solution is
x ≈ 10.61.
70. log4x = log8 4x
log2x — log24
= log24x —
log28
log2x —
2 = log24x
— 3
2(lo g 2 x)/2 = 2(lo g
4 x)/3
x1/2 = (4x)1/3
x3 = (4x)2
x3 = 16x2
x3 − 16x2 = 0 x2(x − 16) = 0 x2 = 0 or x − 16 = 0 x = 0 x = 16
Because log40 is not defi ned, the apparent solution x = 0 is
an extraneous solution. The only solution is x = 16.
71. The equation is quadratic in terms of u = 2x, and can
be factored.
22x − 12 ⋅ 2x + 32 = 0 u2 − 12u + 32 = 0 (u − 8)(u − 4) = 0 (2x − 8)(2x − 4) = 0 2x − 8 = 0 or 2x − 4 = 0 2x = 8 2x = 4 x = 3 x = 2 The solution are x = 3 and x = 2.
72. The equation is quadratic in terms of u = 5x, and can
be factored.
52x + 20 ⋅ 5x − 125 = 0 u2 + 20u − 125 = 0 (u − 5)(u + 25) = 0 (5x − 5)(5x + 25) = 0 5x − 5 = 0 or 5x + 25 = 0 5x = 5 5x = −25
x = 1 Because 5x = −25 is not defi ned, the only solution is x = 1.
332 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
73. To solve exponential equations with different bases, take
a logarithm of each side. Then use the Power Property to
move the exponent to the front of the logarithm, and solve
for x. To solve logarithmic equations of different bases, fi nd
a common multiple of the bases, and exponentiate each side
with this common multiple as the base. Rewrite the base as a
power that will cancel out the given logarithm and solve the
resulting equation.
74. a. I(x) = 0.3I0
I0e−0.43x = 0.3I0
e−0.43x = 0.3
ln e−0.43x = ln 0.3
−0.43x = ln 0.3
x = 1 — −0.43
ln 0.3
x ≈ 2.80
The thickness of the aluminum shielding is about
2.80 centimeters.
b. I(x) = 0.3I0
I0e−3.2x = 0.3I0
e−3.2x = 0.3
ln e−3.2x = ln 0.3
−3.2x = ln 0.3
x = 1 — −3.2
ln 0.3
x ≈ 0.38
The thickness of the copper shielding is about
0.38 centimeter.
c. I(x) = 0.3I0
I0e−43x = 0.3I0
e−43x = 0.3
ln e−43x = ln 0.3
−43x = ln 0.3
x = 1 — −43
ln 0.3
x ≈ 0.03
The thickness of the lead shielding is about 0.03 centimeter.
d. A lead apron does not need to be as thick as aluminum or
copper to result in the same intensity.
Maintaining Mathematical Profi ciency
75. y − y1 = m(x − x1)
y − (−2) = 4(x − 1)
y + 2 = 4x − 4 y = 4x − 6
76. y − y1 = m(x − x1)
y − 2 = −2(x − 3)
y − 2 = −2x + 6 y = −2x + 8
77. y − y1 = m(x − x1)
y − (−8) = − 1 — 3 (x − 3)
y + 8 = − 1 — 3 x + 1
y = − 1 — 3 x − 7
78. y − y1 = m(x − x1)
y − 5 = 2(x − 2)
y − 5 = 2x − 2
79. −50 −13 0
37 13 1 1 13
−24
12 12 12
−12 0 12
12 12
24 36
37 73
1 2 15 52 125
The third differences are constant, so the data can be
modeled by a cubic function. The function is y = 2x3 − x + 1.
80. 139 32 1
−107 −31 −3 1 5
76
−48 −24 0
28
24 24 24 24
4 4
24 48
28 76
33 109
−2 −1 4 37 146
The fourth differences are constant so the data can be
modeled by a quartic function. The function is
y = x4 − 2x3 + x2 + x − 2.
81. −327 −84 −17
243 67 11 3 29
−176
120 48 −24
−56
−72 −72 −72 −72
−8 −32
−96 −168
−128 −296
157 453
−6 −3 −32 −189 −642
The fourth differences are constant so the data can be
modeled by a quartic function. The function is
y = −3x4 + 2x3 − x2 + 5x − 6.
6.7 Explorations (p. 341)
1. a. The type of model matches A. A model for the data is
y = x + 0.5.
b. The type of model matches D. A model for the data is
y = − 1 — 2 x + 5.
c. The type of model matches F. Sample answer: A model
for the data is y = 0.25x2 − 2.
d. The type of model matches C. Sample answer: A model
for the data is y = 0.05x3 − 0.2x2 + 0.25x − 0.4.
e. The type of model matches E. Sample answer: A model
for the data is y = e0.3x.
f. The type of model matches B. Sample answer: A model
for the data is y = log1.4 x.
Copyright © Big Ideas Learning, LLC Algebra 2 333All rights reserved. Worked-Out Solutions
Chapter 6
2. a. y2
x−1−2 21
−1
−2
The domain is all real numbers, the range is 0 < y ≤ 1, the
y-intercept is y = 1, and the asymptote is y = 0.
b. y4
2
x−2−4 42
−2
−4
The domain is all real numbers, the range is 0 < y < 1,
the y-intercept is y = 1 — 2 and the asymptotes are y = 0 and
y = 1.
3. Each of these functions can be characterized by their
domain, range, intercepts, and asymptotes.
4. Sample answer: Hooke’s Law can be modeled by linear data.
x 0 1 2 3 4 5
y 0 2 4 6 8 10
2 4 60 x
F
4
6
8
10
12
20
Distance stretched
Forc
e
Force on a Stretched Spring
The model for this data is F(x) = 2x.
6.7 Monitoring Progress (pp. 342–345)
1. The inputs are equally spaced. Look for a pattern in the
outputs.
x 0 10 20 30
y 15 12 9 6
−3 −3 −3
The fi rst differences are constant. So, the data in the table
represent a linear function.
2. The inputs are equally spaced. Look for a pattern in the
outputs.
x 0 2 4 6
y 27 9 3 1
× 1 —
3 ×
1 —
3 ×
1 —
3
As x increases by 1, y is multiplied by 1 —
3 . So, the constant
ratio is 1 —
3 , and the data in the table represent an exponential
function.
3. Step 1 Substitute the coordinates (2, 12) and (3, 24) into
y = abx.
12 = ab2
24 = ab3
Step 2 Solve for a in Equation 1 to obtain a = 12 —
b2 and
substitute this expression for a in Equation 2.
24 = ( 12 —
b2 ) b3
24 = 12b
2 = b Step 3 Determine that a = 12
— b2 = 12
— 22 = 12
— 4 = 3.
So, the exponential function is y = 3(2x).
4. Step 1 Substitute the coordinates (1, 2) and (3, 32) into
y = abx.
2 = ab1
32 = ab3
Step 2 Solve for a in Equation 1 to obtain a = 2 — b and
substitute this expression for a in Equation 2.
32 = ( 2 — b ) b3
32 = 2b2
16 = b2
4 = b Step 3 Determine that a = 2 —
b = 2 —
4 = 1 —
2 .
So, the exponential function is y = 1 — 2 (4x).
⤻ ⤻ ⤻
334 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
5. Step 1 Substitute the coordinates (2, 16) and (5, 2) into
y = abx.
16 = ab2
2 = ab5
Step 2 Solve for a in Equation 1 to obtain a = 16 —
b2 and
substitute this expression for a in Equation 2.
2 = ( 16 —
b2 ) b5
2 = 16b3
1 —
8 = b3
1 —
2 = b
Step 3 Determine that a = 16
—
( 1 — 2 )
2
= 16
— 1 —
4
= 64.
So, the exponential function is y = 64 ( 1 — 2 )
x
.
6. Sample answer: First fi nd an exponential model using the
points in the table.
Step 1 Make a scatter plot of the data.
x
y
40
80
120
4 62
The data appear exponential.
Step 2 Choose any two points to write a model, such as
(1, 15) and (2, 23). Substitute the coordinates of the
two points into y = abx.
15 = ab1
23 = ab2
Solve for a in the fi rst equation to obtain a = 15 —
b .
Substitute to obtain b = 23 —
15 ≈ 1.53 and a =
15 —
( 23 —
15 ) ≈ 9.78.
So, an exponential function that models the data is
y = 9.78(1.53)x.
Next, fi nd an exponential model using the transformed points
(x, ln y).
Step 1 Create a table of data pairs (x, ln y).
x 1 2 3 4 5 6 7
ln y 2.71 3.14 3.69 3.95 4.38 4.65 4.94
Step 2 Plot the transformed points as shown.
x
ln y
2
4
6
4 62
The point lie close to a line, so an exponential model
should be a good fi t for the original data.
Step 3 Find an exponential model y = abx by choosing any
two points on the line, such as (1, 2.71) and (2, 3.14).
Use these points to write an equation of the line.
Then solve for y.
ln y − 2.71 = 0.43(x − 1)
ln y = 0.43x + 2.28
y = e0.43x+2.28
y = e0.43x(e2.28)
y = 9.77(1.54)x
So, an exponential function that models the data is
y = 9.77(1.54)x.
7. Enter the data into a graphing calculator and perform an
exponential regression. The model is y = 11.39(1.45)x.
8. The data do not model a logarithmic function.
6.7 Exercises (pp. 346–348)
Vocabulary and Core Concept Check
1. Given a set of more than two data pairs (x, y), you can decide
whether an exponential function fi ts the data well by making
a scatter plot of the points (x, ln y).
2. If the inputs are equally spaced and the outputs are
multiplied by a constant factor, then the data can be
represented by an exponential function model.
Monitoring Progress and Modeling with Mathematics
3. The inputs are equally spaced. Look for a pattern in the
outputs.
x 0 3 6 9 12 15
y 0.25 1 4 16 64 256
×4 ×4 ×4 ×4 ×4
As x increases by 3, y is multiplied by 4. So, the constant
ratio is 4, and the data in the table represent an exponential
function.
⤻ ⤻⤻ ⤻⤻
Copyright © Big Ideas Learning, LLC Algebra 2 335All rights reserved. Worked-Out Solutions
Chapter 6
4. The inputs are equally spaced. Look for a pattern in the
outputs.
x −4 −3 −2 −1 0 1 2
y 16 8 4 2 1 1 —
2
1 —
4
× 1 — 2 × 1 —
2 × 1 —
2 ×
1 —
2 × 1 —
2 × 1 —
2
As x increases by 1, y is multiplied by 1 —
2 . So, the constant
ratio is 1 —
2 , and the data in the table represent an exponential
function.
5. The inputs are equally spaced. The outputs do not have a
common ratio. So, analyze the fi nite differences.
x 5 10 15 20 25 30
y 4 3 7 16 30 49
−1 4 9 14 19
5 5 5 5
The second differences are constant. So, the data in the table
represent a quadratic function.
6. The inputs are equally spaced. The outputs do not have a
common ratio. So, analyze the fi nite differences.
x −3 1 5 9 13
y 8 −3 −14 −25 −36
−11 −11 −11 −11
The fi rst differences are constant. So, the data in the table
represent a linear function.
7. Step 1 Substitute the coordinates (1, 3) and (2, 12) into
y = abx.
3 = ab1
12 = ab2
Step 2 Solve for a in Equation 1 to obtain a = 3 — b and
substitute this expression for a in Equation 2.
12 = ( 3 — b ) b2
12 = 3b
4 = b Step 3 Determine that a = 3 —
b = 3 —
4 .
So, the exponential function is y = 3 — 4 (4)x.
8. Step 1 Substitute the coordinates (2, 24) and (3, 144) into
y = abx.
24 = ab2
144 = ab3
Step 2 Solve for a in Equation 1 to obtain a = 24 —
b2 and
substitute this expression for a into Equation 2.
144 = ( 24 —
b2 ) b3
144 = 24b
6 = b Step 3 Determine that a = 24
— b = 24
— 36
= 2 — 3 .
So, the exponential function is y = 2 — 3 (6)x.
9. Step 1 Substitute the coordinates (3,1) and (5, 4) into
y = abx.
1 = ab3
4 = ab5
Step 2 Solve for a in Equation 1 to obtain a = 1 — b3 and
substitute this expression for a into Equation 2.
4 = ( 1 — b3 ) b5
4 = b2
2 = b Step 3 Determine that a = 1 —
b3 = 1 — 23 = 1 —
8 .
So, the exponential function is y = 1 — 8 (2)x.
10. Step 1 Substitute the coordinates (3, 27) and (5, 243) into
y = abx.
27 = ab3
243 = ab5
Step 2 Solve for a in Equation 1 to obtain a = 27 —
b3 and
substitute this expression for a into Equation 2.
243 = ( 27 —
b3 ) b5
243 = 27b2
9 = b2
3 = b Step 3 Determine that a = 27
— b3 = 27
— 33 = 1.
So, the exponential function is y = 1 ⋅ (3x) = 3x.
⤻⤻ ⤻⤻ ⤻⤻
336 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
11. Step 1 Substitute the coordinates (1, 2) and (3, 50) into
y = abx.
2 = ab1
50 = ab3
Step 2 Solve for a in Equation 1 to obtain a = 2 — b and
substitute this expression for a into Equation 2.
50 = ( 2 — b ) b3
50 = 2b2
25 = b2
5 = b Step 3 Determine that a = 2 —
b = 2 —
5 .
So, the exponential function is y = 2 — 5 (5)x.
12. Step 1 Substitute the coordinates (1, 40) and (3, 640) into
y = abx.
40 = ab1
640 = ab3
Step 2 Solve for a in Equation 1 to obtain a = 40 —
b and
substitute this expression for a into Equation 2.
640 = ( 40 —
b ) b3
640 = 40b2
16 = b2
4 = b Step 3 Determine that a = 40
— b = 40
— 4 = 10.
So, the exponential function is y = 10(4)x.
13. Step 1 Substitute the coordinates (−1, 10) and (4, 0.31) into
y = abx.
10 = ab−1
0.31 = ab4
Step 2 Solve for a in Equation 1 to obtain a = 10b and
substitute this expression for a into Equation 2.
0.31 = (10b)b4
0.31 = 10b5
0.031 = b5
0.5 ≈ b Step 3 Determine that a ≈ 10(0.5) = 5.
So, the exponential function is y = 5(0.5)x.
14. Step 1 Substitute the coordinates (2, 6.4) and (5, 409.6) into
y = abx.
6.4 = ab2
409.6 = ab5
Step 2 Solve for a in Equation 1 to obtain a = 64 —
b2 and
substitute this expression for a into Equation 2.
409.6 = ( 6.4 —
b2 ) b5
409.6 = 6.4b3
64 = b3
4 = b Step 3 Determine that a = 6.4
— b2 = 6.4
— 42 = 0.4.
So, the exponential function is y = 0.4(4)x.
15. Step 1 Substitute the coordinates (1, 0.5) and (4, 4) into
y = abx.
0.5 = ab1
4 = ab4
Step 2 Solve for a in Equation 1 to obtain a = 0.5 —
b and
substitute this expression for a into Equation 2.
4 = ( 0.5 —
b ) b4
4 = 0.5b3
8 = b3
2 = b Step 3 Determine that a = 0.5
— b = 0.5
— 2 = 0.25.
So, the exponential function is y = 0.25(2)x.
16. Step 1 Substitute the coordinates (−3, 10.8) and (−2, 3.6)
into y = abx.
3.6 = ab−2
10.8 = ab−3
Step 2 Solve for a in Equation 1 to obtain a = 3.6b2 and
substitute this expression for a in Equation 2.
10.8 = (3.6b2)b−3
10.8 = 3.6b−1
3 = b−1
1 —
3 = b
Step 3 Determine that a = 3.6 ( 1 — 3 )
2
= 0.4.
So, the exponential function is y = 0.4 ( 1 — 3 )
x
.
17. Data are linear when the fi rst differences are constant.
Because there is a constant ratio of 3, the data represent an
exponential function.
18. The error is that the inputs are not equally spaced. The data
are modeled by a quartic function that can be found using the
regression feature of a graphing calculator.
Copyright © Big Ideas Learning, LLC Algebra 2 337All rights reserved. Worked-Out Solutions
Chapter 6
19. Sample answer:
Step 1 Make a scatter plot of the data.
x
y
20
40
60
80
4 62
The data appear exponential.
Step 2 Choose any two points to write a model, such as
(1, 9) and (2, 14). Substitute the coordinates of these
two points into y = abx.
9 = ab1
14 = ab2
Solve for a in the fi rst equation to obtain a = 9 — b .
Substitute to obtain b = 14 —
9 ≈ 1.56 and
a = 9
— 14
— 9
= 81 —
14 ≈ 5.79.
So, an exponential function that represents the data is
y = 5.79(1.56x).
20. Sample answer:
Step 1 Make a scatter plot of the data.
x
y
200
400
600
800
4 62
The data appear exponential.
Step 2 Choose any two points to write a model, such as
(1, 22) and (2, 39). Substitute the coordinates of
these points into y = ab x.
22 = ab1
39 = ab2
Solve for a in the fi rst equation to obtain a = 22 —
b .
Substitute to obtain b = 39 —
22 ≈ 1.78 and
a =
22
— 39
— 22
≈ 12.41.
So, an exponential function that models the data is y = 12.41(1.78x). When the time is 1 year, the number of
predicted visits is y = 12.41(1.7812) ≈ 12,555.
21. Sample answer: Make a scatter plot of the data.
x
y
120
240
360
480
12 186
The data appear exponential. Choose any two points to
write a model, such as (1, 12) and (6, 28). Substitute the
coordinates of these points into y = ab x.
12 = ab1
28 = ab6
Solve for a in the fi rst equation to obtain a = 12 —
b . Substitute
to obtain b ≈ 1.19 and a ≈ 10.13. So, an exponential
function that models the data is y = 10.13(1.19 x).
22. Sample answer: Make a scatter plot of the data.
x
y
50
100
150
200
126−6
The data appear exponential. Choose any two points to
write a model, such as (1, 24) and (3, 68). Substitute the
coordinates of these points into y = ab x.
24 = ab1
68 = ab3
Solve for a in the fi rst equation to obtain a = 24
— b . Substitute
to obtain b ≈ 1.68 and a ≈ 14.29. So, an exponential
function that models the data is y = 14.29(1.68x).
338 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
23. Sample answer: Make a scatter plot of the data.
x
y
20
40
60
80
40 6020
The data do not appear exponential. Choose any two points
to write a linear model, such as (30, 42) and (50, 26). Find
the slope and use the point-slope form to write an equation
of the model.
m = y2 − y1 — x2 − x1
= 26 − 42
— 50 − 30
= −16
— 20
= −0.8
y − y1 = m(x − x1)
y − 42 = −0.8(x − 30)
y − 42 = −0.8x + 24
y = −0.8x + 66
A linear model for the data is y = −0.8x + 66.
24. Sample answer: Make a scatter plot of the data.
x
y
10
20
30
40
2010−10−20
The data appear exponential. Choose any two points to write
a model, such as (1, 11) and (8, 8). Substitute the coordinates
of these points into y = abx.
11 = ab1
8 = ab8
Solve for a in the fi rst equation to obtain a = 11
— b . Substitute
to obtain b ≈ 0.96 and a ≈ 11.51. So, an exponential
function that models the data is y = 11.51(0.96)x.
25. Step 1 Create a table of data pairs (x, ln y).
x 20 30 40 50 60
ln y 2.48 2.71 3.22 3.69 4.61
Step 2 Plot the transformed points.
ln y
40 60200
6
2
0
4
x
The points lie close to a line, so an exponential
model should be a good fi t for the original data.
Step 3 Find an exponential model y = ab x by choosing any
two points, such as (20, 2.48) and (30, 3.22). Use
these points to write an equation of the line. Then
solve for y.
The slope is 3.22 − 2.48
— 30 − 20
≈ 0.74
— 10
≈ 0.7. Because the
axes are x and ln y, the point-slope form is rewritten
as ln y − ln y1 = m(x − x1).
ln y − 2.48 = 0.07(x − 20)
ln y = 0.07x + 1.08
y = e0.07x+1.08
y = e0.07x(e1.08)
y = 2.94(1.07) x
So, an exponential function that models the data is
y = 2.94(1.07)x.
Copyright © Big Ideas Learning, LLC Algebra 2 339All rights reserved. Worked-Out Solutions
Chapter 6
26. Step 1 Create a table of data pairs (x, ln y).
x 1 2 3 4 5 6 7
ln y 2.20 2.64 2.94 3.22 3.61 3.97 4.26
Step 2 Plot the transformed points.
ln y
4 620
6
2
0
4
x
The points lie close to a line, so an exponential model
should be a good fi t for the original data.
Step 3 Find an exponential model y = ab x by choosing any
two points, such as (1, 2.20) and (2, 2.64). Use these
points to write an equation of the line. Then solve for y.
The slope is 2.64 − 2.20
— 2− 1
= 0.44
— 1 = 0.44.
ln y − ln y1 = m(x − x1)
ln y − 2.20 = 0.44(x − 1)
ln y = 0.44x + 1.76
y = e0.44x+1.76
y = e0.44x(e1.76)
y = 5.81(1.55)x
So, an exponential function that models the data is
y = 5.81(1.55)x.
27. Create a table of data pairs (x, ln y).
x 1 2 3 4 5
ln y 2.89 3.58 5.92 4.97 5.66
Make a scatter plot of the points in the table.
ln y
4 620
6
2
0
4
x
The points appear linear.
Find an exponential model y = ab x by choosing two points,
such as (1, 2.89) and (2, 3.58). Use these points to write an
equation of the line. Then solve for y.
The slope is 3.58 − 2.89
— 2− 1
= 0.69
— 1 = 0.69.
ln y − ln y1 = m(x − x1)
ln y − 2.89 = 0.69(x − 1)
ln y = 0.69x + 2.2
y = e0.69x+2.2
y = e0.69x(e2.2)
y = 9.03(1.99)x
So, an exponential function that models the data is
y = 9.03(1.99)x.
28. Create a table of data pairs (x, ln y).
x 1 4 7 10 13
ln y 1.19 2.31 3.42 4.53 5.64
Make a scatter plot of the points in the table.
ln y
4 6 8 10 1220
6
2
0
4
x
The points appear linear.
Find an exponential model y = ab x by choosing two points,
such as (1, 1.19) and (4, 2.31). Use these points to write an
equation of the line. Then solve for y.
The slope is 2.31 − 1.19
— 4 − 1
= 1.12
— 3 ≈ 0.37.
ln y − ln y1 = m(x − x1)
ln y − 1.19 = 0.37(x − 1)
ln y = 0.37x + 0.82
y = e0.37x+0.82
y = e0.37x(e0.82)
y = 2.27(1.44)x.
So, an exponential function that models the data is
y = 2.27(1.44) x.
29. Create a table of data pairs (x, ln y).
x −13 −6 1 8 15
ln y 2.28 2.50 2.72 2.94 3.17
Make a scatter plot of the points in the table.
8 124−4−8
4
2
x
ln y
−12
The points appear linear.
Find an exponential model y = ab x by choosing two points,
such as (−13, 2.28) and (1, 2.72). Use these points to write
an equation of the line. Then solve for y.
The slope is 2.72 − 2.28
— 1 − (−13)
= 0.44
— 14
≈ 0.03.
ln y − ln y1 = m(x − x1)
ln y − 2.72 = 0.03(x − 1)
ln y = 0.03x + 2.69
y = e0.03x+2.69
y = e0.03x(e2.69)
y = 14.73(1.03)x
So, an exponential function that models the data is
y = 14.73(1.03)x.
340 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
30. Create a table of data pairs (x, ln y).
x −8 −5 −2 1 4
ln y 0.33 0.51 1.67 1.86 2.08
Make a scatter plot of the points in the table.
42−2−4−6−8
2
x
ln y
The points do not appear linear, so an exponential model
does not fi t the original data.
31. Enter the data into a graphing calculator and perform an
exponential regression. The model is y = 6.7(1.41)x. In the
tenth year, the number of scooters sold is expected to be
about y = 6.7(1.41)10 ≈ 208.
32. Enter the data into a graphing calculator and perform an
exponential regression. The model is y = 153.07(0.93)x.
After 16 minutes the astronaut’s pulse rate is expected to be
about y = 153.07(0.93)16 ≈ 48 beats per minute.
33. Enter the data into a graphing calculator and perform a
logarithmic regression. The model is t = 12.59 − 2.55 ln d.
It takes about t = 12.59 − 2.55 ln 50 ≈ 2.6 hours for the
object to cool to 50° C.
34. Enter the data into a graphing calculator and perform a
logarithmic regression. The model is
s = 0.000398 + 2.89 ln f. The amount of light that strikes
the camera when f = 5.657 is about
s = 0.000398 + 2.89 ln 5.657 ≈ 5 units of light.
35. a. Sample answer: Make a scatter plot of the points (x, ln y).
2 3 4 51−1
0.5
1
1.5
x
ln y
−0.5(1, −0.29)
(2, 0.08)
(3, 0.53)(4, 0.79)
(5, 1.23)
The points appear linear, so enter the points (x, y)
into a graphing calculator and perform an exponential
regression. The model is y = 0.52(1.46)x.
b. Because the growth factor is 1.46, the weight increases by
0.46 or 46% per year.
36. The points (x, y) fi t an exponential pattern because the points
(x, ln y) fi t a linear pattern.
37. Your friend is incorrect because t = 0 is not in the domain of
ln t.
38. It is possible to write an exponential function because the
ratios of the outputs are equal to 2.
39. a. 200 = 256 ——
1 + 13e−0.65t
200(1 + 13e−0.65t) = 256
1 + 13e−0.65t = 1.28
13e−0.65t = 0.28
e−0.65t ≈ 0.022
−0.65t ≈ ln 0.022
t ≈ 1 —
−0.65 ln 0.022
t ≈ 5.9
It takes about 5.9 weeks for the sunfl ower seedling to
reach 200 centimeters.
b. y
4 6 8 10 12 1420
300
100
0
200
x
The asymptote is the line y = 256 and represents the
maximum height of the sunfl ower.
Maintaining Mathematical Profi ciency
40. The variables x and y have a proportional relationship
because the equation has the form y = kx.
41. The variables x and y do not have a proportional relationship
because the equation does not have the form y = kx.
42. The variables x and y do not have a proportional relationship
because the equation does not have the form y = kx.
43. The variables x and y have a proportional relationship
because the equation has the form y = kx.
44. The equation has the form x = 1 —
4p y2, where p = 2. The
focus is (p, 0), or (2, 0). The directrix is x = −p, or x = −2.
Because y is squared, the axis of symmetry is y = 0.
y
8 12 164−4
8
4
x
−4
−8
−12
12
Copyright © Big Ideas Learning, LLC Algebra 2 341All rights reserved. Worked-Out Solutions
Chapter 6
45. The equation has the form y = 1 —
4p x2, where p =
1 —
16 .
The focus is (0, p), or ( 0, 1 —
16 ) . The directrix is y = −p,
or y = − 1 —
16 . Because x is squared, the axis of symmetry
is x = 0.
y
42−2−4
4
6
2
x
46. The equation has the form y = 1 —
4p x2, where p =
3 —
4 . The
focus is (0, p), or ( 0, 3 —
4 ) . The directrix is y = −p, or y = −
3 —
4 .
Because x is squared, the axis of symmetry is x = 0.
y
42−2−4
4
6
8
2
x
47. The equation has the form x = 1 —
4p y2, where p =
1 —
10 . The
focus is ( p, 0), or ( 1 — 10
, 0 ) . The directrix is x = −p, or
x = − 1 — 10
. Because y is squared, the axis of symmetry is
y = 0.
y
4 62
4
2
x
−2
−4
6.5–6.7 What Did You Learn? (p. 349)
1. The change-of-base formula is used to change the base from
e to 10, then the expression is rewritten using properties of
logarithms.
2. Any case where a logarithmic function has a solution that is
negative disproves the theory. From that, it is easy to create
an expression that becomes positive when a negative value is
substituted for x and create a logarithmic equation with it.
Chapter 6 Review (pp. 350–352)
1. The base, 1 —
3 , is greater than 0 and less than 1, so the function
represents exponential decay. Because the decay factor is
1 —
3 = 1 −
2 —
3 , the percent decrease is
2 —
3 , or about 66.67%.
y
42−2−4
4
6
8
x
2. The base, 5, is greater than 1, so the function represents
exponential growth. Because the growth factor is 5 = 1 + 4,
the percent increase is 4, or 400%.
y
42−2−4
2
4
6
8
x
3. The base, 0.2, is greater than 0 and less than 1, so the
function represents exponential decay. Because the decay
factor is 0.2 = 1 − 0.8, the percent decrease is 0.8, or 80%.
y
42−2−4
4
6
8
x
4. The interest is compounded daily, so n = 365. The balance
after 2 years is
A = P ( 1 + r —
n )
n⋅t
= 1500 ( 1 + 0.07
— 365
) 365⋅2
≈ 1725.39.
The balance at the end of 2 years is $1725.39.
5. e4 ⋅ e11 = e4+11
= e15
342 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
6. 20e3 —
10e6 = 2e3−6
= 2e−3
= 2 —
e3
7. (−3e−5x)2 = (−3)2(e−5x)2
= 9e2(−5x)
= 9e−10x
= 9 —
e10x
8. Because a = 1 —
3 and r = 1 is positive, the function is an
exponential growth function. Use a table to graph the
function.
x −2 −1 0 1
f (x) 0.05 0.12 0.33 0.91
y
42−2−4
2
4
6
8
x
9. Because a = 6 and r = −1 is negative, the function is an
exponential decay function. Use a table to graph the function.
x −2 −1 0 1
y 44.33 16.31 6 2.21
y
42−2−4
4
x
16
10. Because a = 3 and r = −0.75 is negative, the function is an
exponential decay function. Use a table to graph the function.
x −2 −1 0 1
y 13.45 6.35 3 1.42
y
42−2−4
8
x
16
12
11. Because 23 = 8, log2 8 = 3.
12. Because 6−2 = 1 —
36 , log6
1 —
36 = −2.
13. Because 50 = 1, log5 1 = 0.
14. From the defi nition of logarithm, the inverse of f (x) = 8x is
g(x) = log8 x.
15. y = ln(x − 4)
x = ln(y − 4)
ex = eln(y−4)
ex = y − 4
ex + 4 = y
The inverse of y = ln(x − 4) is y = ex + 4.
16. y = log(x + 9)
x = log(y + 9)
10 x = 10log(y+9)
10 x = y + 9
10 x − 9 = y
The inverse of y = log(x + 9) is y = 10 x − 9.
17. Step 1 Find the inverse of y = log1�5 x. From the defi nition
of logarithm, the inverse of y = log1�5 x is y = ( 1 — 5 )
x
.
Step 2 Make a table of values for y = ( 1 — 5 )
x
.
x −2 −1 0 1 2
y 25 5 1 1 —
5
1 —
25
Copyright © Big Ideas Learning, LLC Algebra 2 343All rights reserved. Worked-Out Solutions
Chapter 6
Step 3 Because y = log1�5 x and y = ( 1 — 5 )
x
are inverse
functions, the graph of y = log1�5 x is obtained by
refl ecting the graph of y = ( 1 — 5 )
x
in the line y = x.
To do this, reverse the coordinates of the points of
y = ( 1 — 5 )
x
and plot these new points to obtain the
graph of y = log1�5 x.
y
4 6
2
4
x
−2
18. Notice that the function is of the form g(x) = e−ax + k,
where a = 5 and k = −8. So, the graph of g is a horizontal
shrink by a factor of 1 —
5 followed by a translation 8 units down
of the graph of f.
y
42
4
8
x−4 −2
−4
−8g
f
19. Notice that the function is of the form g(x) = a log4(x − h),
where h = −5 and a = 1 —
2 . So, the graph of g is a vertical
shrink by a factor of 1 —
2 and a translation 5 units left of the
graph of f.
y
4 62
2
4
x−4 −2
−2
−4
g f
20. Step 1 First write a function h that represents the vertical
stretch of f.
h(x) = 3f (x)
= 3e x
Step 2 Then write a function g that represents the translation
of h.
g(x) = h(x + 6) + 3
= 3e x+6 + 3
The transformed function is g(x) = 3e x+6 + 3.
21. Step 1 First write a function h that represents the translation
of f.
h(x) = f (x) − 2
= log x − 2
Step 2 Then write a function g that represents the refl ection
of h.
g(x) = h(−x)
= log(−x) − 2
The transformed function is g(x) = log (−x) − 2.
22. log8 3xy = log8 3 + log8 x + log8 y
23. log 10x3y = log 10 + log x3 + log y
= 1 + 3 log x + log y
24. ln 3y
— x5
= ln 3y − ln x5
= ln 3 + ln y − 5 ln x
25. 3 log7 4 + log7 6 = log7 43 + log7 6
= log7 (43 ⋅ 6)
= log7 384
26. log2 12 − 2 log2 x = log2 12 − log2 x2
= log2 12
— x2
27. 2 ln x + 5 ln 2 − ln 8 = ln x2 + ln 25 − ln 8
= ln 32x2
— 8
= ln 4x2
28. log2 10 = log 10
— log 2
≈ 3.32
29. log7 9 = log 9
— log 7
≈ 1.13
30. log23 42 = log 42
— log 23
≈ 1.19
31. 5x = 8
log5 5x = log5 8
x = log5 8
x ≈ 1.29
The solution is x = log5 8, or about 1.29.
32. log3 (2x − 5) = 2 Check
3 log 3 (2x−5) = 32 log3 (2 ⋅ 7 − 5) =
? 2
2x − 5 = 9 log3 9 =?
2 2x = 14 2 = 2 ✓
x = 7 The solution is x = 7.
344 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
33. ln x + ln (x + 2) = 3
ln [ x(x + 2) ] = 3
eln[x(x+2)] = e3
x(x + 2) = e3
x2 + 2x = e3
x2 + 2x − e3 = 0
x ≈ −5.59 or x ≈ −3.59
Check
ln (−5.59) + ln (−5.59 + 2) =?
3 Because ln (−5.59) is not defi ned, −5.59 is not a solution.
ln 3.59 + ln (3.59 + 2) =?
3 ln 3.59 + ln 5.59 =
? 3
1.278 + 1.721 ≈ 3 ✓
The apparent solution x ≈ −5.59 is extraneous. The only
solution is x ≈ 3.59.
34. 6x > 12
log6 6x > log6 12
x > log6 12
The solution is x > log6 12. Because log6 12 ≈ 1.39, the
approximate solution is x > 1.39.
35. ln x ≤ 9
eln x ≤ e9
x ≤ e9
The solution is x ≤ e9. Because ln x is only defi ned
when x > 0 and e9 ≈ 8103, the approximate solution
is 0 < x ≤ 8103.
36. e4x−2 ≥ 16
ln (e4x−2) ≥ ln 16
4x − 2 ≥ ln 16
4x ≥ ln 16 + 2
x ≥ 1 —
4 ln 16 + 1 —
2
The solution is x ≥ 1 —
4 ln 16 + 1 —
2 . Because
1 —
4 ln 16 + 1 —
2 ≈ 1.19, the approximate solution is x ≥ 1.19.
37. Step 1 Substitute the coordinates of the two given points
into y = abx.
8 = ab3
2 = ab5
Step 2 Solve for a in Equation 1 to obtain a = 8 — b3 and
substitute this expression for a in Equation 2.
2 = ( 8 — b3 ) b5
2 = 8b2
1 —
4 = b2
1 —
2 = b
Step 3 Determine that a = 8 — b3 = 8
____
( 1 — 2 )
3
= 64.
So, the exponential function is y = 64 ( 1 — 2 )
x
.
38. Find an exponential model y = abx by choosing any two
points, such as (1, 1.64) and (2, 2.00). Use these points to
write an equation of the line through the points (x, ln y).
Then solve for y.
The slope is 2.00 − 1.64
— 2− 1
= 0.36
— 1 = 0.36.
ln y − ln y1 = m(x − x1)
ln y − 1.64 = 0.36(x − 1)
ln y = 0.36x + 1.28
y = e0.36x+1.28
y = e0.36x(e1.28)
y ≈ 3.60(1.43)x
So, an exponential function that models the data is
y = 3.60(1.43)x.
39. Enter the data into a graphing calculator and perform a
logarithmic regression. The model is s = 3.95 + 27.48 ln t. There are about s = 3.95 + 27.48 ln 6 ≈ 53 pairs of shoes
sold after week 6.
Chapter 6 Test (p. 353)
1. y
42−2−4
4
2
6
8
x
The domain is all real
numbers, the range is
y > 0, and the asymptote
is y = 0.
Copyright © Big Ideas Learning, LLC Algebra 2 345All rights reserved. Worked-Out Solutions
Chapter 6
2. y
4 6 8
4
2
6
x
−2
The domain is x > 0, the
range is all real numbers,
and the asymptote is x = 0.
3. y
42−2−4
4
x
12
16
The domain is all real
numbers, the range is
y > 0, and the asymptote
is y = 0.
4. The graph of g is a refl ection in the x-axis followed by
a translation 4 units right of the graph of f. So,
g(x) = −log(x − 4).
5. The graph of g is a refl ection in the y-axis followed by a
translation 2 units up of the graph of f. So, g(x) = e−x + 2.
6. The graph of g is a vertical stretch by a factor of 2 of the
graph of f . So, g(x) = 2 ( 1 — 4 )
x
.
7. log3 52 = log3 4 + log3 13
≈ 1.262 + 2.335
= 3.597
8. log3 13
— 9 = log3 13 − log3 9
= log3 13 − 2 ≈ 2.335 − 2 = 0.335
9. log3 16 = 2 log3 4
≈ 2(1.262)
= 2.524
10. log3 8 + log3 1 —
2 = log3 2 + log3 4 − log3 2
= log3 4
≈ 1.262
11. To solve the fi rst equation use a logarithm on both sides and
to solve the second equation use exponentials on both sides.
45x−2 = 16 log4(10x + 6) = 1 log4 4
5x−2 = log4 16 4 log 4 (10x+6) = 41
5x − 2 = 2 10x + 6 = 4 5x = 4 10x = −2
x = 4 — 5 x = − 1 —
5
12. All three are equivalent by the change-of-base formula.
13. One thousand wells will produce
y = 12.263 ln 1000 − 45.381 ≈ 39 billion barrels.
y = 12.263 ln x − 45.381
x = 12.263 ln y − 45.381
x + 45.381 = 12.263 ln y
x + 45.381
— 12.263
= ln y
e x+45.381
— 12.263
= y
e(x�12.263) + (45.381/12.263) = y ( ex�12.263 ) ( e45.381�12.263 ) = y ( e1�12.263 )
x ( e45.381�12.263 ) = y
40.473(1.085)x ≈ y The function y ≈ 40.473(1.085)x represents the number
of wells needed to produce a certain number of billions
of barrels of oil.
14. a. The function is L(x) = 100e−0.02x.
b. The function in part (a) represents exponential decay since
the base, e−0.02 ≈ 0.98, is greater than 0 and less than 1.
c. When the depth is 40 meters, the percent of surface light
is L(40) = 100 e−0.02(40) ≈ 44.9 or about 44.9%.
15. Sample answer: The three ways to fi nd the exponential
model are:
1. Use two points and the model y = abx to determine the
values of a and b.
2. Convert the pairs to (x, ln y), then solve the related linear
equation for y.
3. Enter the points into a graphing calculator and perform
exponential regression.
The model is y = 4200 (0.89)x and the snowmobile is worth
$2500 in about 4.5 years.
Chapter 6 Standards Assessment (pp. 354–355)
1. The possible values for b are 0.94 and e−1/2.
2. Your friend’s claim is correct, interest compounded
continuously produces the most interest when compound
interest is used.
346 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 6
3. a. T(x) = (x)(4x)(2x) = 8x3
b. C(x) = (x − 2)(4x − 2)(2x − 4) = 8x3 − 36x2 + 48x − 16
c. The relationship is I(x) = T(x) − C(x).
d. I(x) = T(x) − C(x)
= (8x3) − (8x3 − 36x2 + 48x − 16)
= 36x2 − 48x + 16
The volume of the insulation is I(8) = 36(8)2 − 48(8) + 16 = 1936 cubic inches when the width is 8 inches.
4. −4 log2 x ≥ −20
log2 x ≤ 5
2 log2 x ≤ 25
x ≤ 32
Because log2 x is only defi ned when x > 0, the solution is
0 < x ≤ 32, which is choice C.
5. The graph of g is a refl ection in the y-axis followed by a
translation 2 units down of the graph of f.
6. (f + g)(x) = f (x) + g(x) = −3x4 − 4x3 − 4x2 + 10x + 4 (hg)(x) = h(x) g(x) = −3x6 − 9x5 − 15x4 + 44x3 + 7x2 − 9x − 35
(h − f )(x) = h(x) − f (x) = −2x3 + 5x2 − 7x − 6 ( f h)(x) = f (x) h(x) = 2x5 − 2x4 − 10x3 − 35x2 − 57x + 7 The order of the polynomials from least degree to greatest
degree is C, A, D, B.
7. a. Substitute the coordinates of two points, such as (2, 20)
and (3, 40), into the model y = abx.
20 = ab2
40 = ab3
Solve for a in the fi rst equation to obtain a = 20 —
b2 , then
b = 2 and a = 5. So, the exponential model is y = 5(2)x.
b. Substitute the coordinates of two points, such as (2, 4.5)
and (3, 13.5), into the model y = abx.
4.5 = ab2
13.5 = ab3
Solve for a in the fi rst equation to obtain a = 4.5 —
b2 , then
b = 3 and a = 1 —
2 . So, the exponential model is y = 1 —
2 (3)x.
The equation in part (a) has a larger y-intercept and grows
at a rate of 100%. The equation in part (b) has a smaller
y-intercept but grows at a faster rate of 200% and will
become larger than the equation in part (a) for x > 5.7.
8. a. Sample answer: Use the Quadratic Formula because the
polynomial x2 + 4x − 10 does not factor.
b. Sample answer: Use the Square Root method because
there is only one term with a variable.
c. Sample answer: Use the Quadratic Formula because the
resulting polynomial does not factor.
d. Sample answer: Use the factoring method because
x2 − 3x − 18 factors.
9. y
40 60200
300
400
500
100
0
200
x
The scatter plot shows a quadratic relationship. Enter the
data into a graphing calculator and perform quadratic
regression. The model is y = −0.261x2 + 22.591x + 23.029.
When y = 500, the model estimates x to be about 50.0°
or 36.5°.