Chapter 6whssalisbury.weebly.com/uploads/1/1/2/8/112801805/hscc_alg2_ws… · Chapter 6 Chapter 6...

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.


x −3 −2 −1 0 1 2

y 27

— 8

9 —

4

3 —

2 1

2 —

3

4 —

9


Step 4 Draw, from right to left, a smooth curve that begins


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.


x −1.5 −1 −0.5 0 0.5 1

y 8 4 2 1 0.5 0.25





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.


x −2 −1 0 1 2 3

y 4 —

9

2 —

3 1

3 —

2

9 —

4

27 —

8



Chapter 6




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




x −2 −1 0 1

y 1 —

36

1 —

6 1 6



above the x-axis, passes through the plotted points,

and moves up to the right.

y

42−2−4

2

4

6

8

x


Chapter 6




x −2 −1 0 1

y 1 —

49

1 —

7 1 7



above the x-axis, passes through the plotted points,

and moves up to the right.

y

42−2−4

2

4

6

8

x


6 , is greater


exponential decay.


x −1 0 1 2

y 6 1 1 —

6

1 —

36





y

42−2−4

4

6

8

x


8 , is greater


exponential decay.


x −1 0 1 2

y 8 1 1 —

8

1 —

64





y

42−2−4

6

8

x


3 , is greater



x −2 −1 0 1 2 3

y 9 —

16

3 —

4 1

4 —

3

16 —

9

64 —

27





y

4 6 82−2

2

4

6

8

x

10


Chapter 6


5 , is greater


exponential decay.


x −2 −1 0 1 2

y 25

— 4

5 —

2 1

2 —

5

4 —

25





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.


x −2 −1 0 1 2 3

y 0.69 — 4 0.8

— 3 1 1.2 1.44 1.728





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



x −3 −2 −1 0 1 2

y 2. — 370 1.

— 7 1. — 3 1 0.75 0.5625





y

42−2−4−6−8

2

4

6

8

x

10

12


greater than 0 and less than 1, so the function



x −2 −1 0 1 2 3

y 2. — 7 1.

— 6 1 0.6 0.36 0.216





y

42−2−4

2

4

6

8

x


Chapter 6


greater than 1, so the function represents exponential

growth.


x −1 0 1 2 3

y 0. — 5 1 1.8 3.24 5.832




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%.


Chapter 6

30. y = a(4)t/6

= a[(4)1/6]t

≈ a(1.26)t

= a(1 + 0.26)t


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


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.


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.


in interest.


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


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


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


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.


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.


x −2 −1 0 1

y 29.56 10.87 4 1.47

y

42−2−4

2

4

8

x


Chapter 6

6. Because a = 2 is positive and r = 2 is positive, 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.



1. an irrational number that is approximately 2.718281828

2. Because a = 1 — 3 is positive and r = 4 is positive, the function



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




x −2 −1 0 1

y 0.002 0.05 1 20.09

y

42−2−4

2

4

6

8

x


Chapter 6




x −2 −1 0 1

y 54.6 7.39 1 0.14

y

42−2−4

4

6

8

x




x −2 −1 0 1

y 14.78 5.44 2 0.74

y

42−2−4

2

4

6

8

x




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.


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



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



x −2 −1 0 1

y 0.66 0.51 0.4 0.31

y

4−4−8

4

8

x

16

12

−12


Chapter 6

22. Because a = 0.6 and r = 0.5 are positive, the function is an

exponential growth 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



33. x −2 −1 0 1

y 1.27 1.74 3 6.44

y

42−2−4

8

4

x

16

12




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


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.


44. 0.006 = 6 × 10−3

45. 5000 = 5 × 103

46. 26,000,000 = 2.6 × 107

47. 0.000000047 = 4.7 × 10−8


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——


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——


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.


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


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


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


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)


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.


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


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.


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


Step 2 Make a table of values for y = 4x.

x −2 −1 0 1 2

y 1 —

16

1 —

4 1 4 16


a smooth curve.


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


Step 2 Make a table of values of y = 6x.

x −2 −1 0 1 2

y 1 —

36

1 —

6 1 6 36


a smooth curve.


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


a smooth curve.


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


y

4 6 8 10

2

4

6

x

−2

−4


Chapter 6



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


a smooth curve.


x

are inverse



x

in the line y = x.


y = ( 1 — 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


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


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


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


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


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.


Chapter 6

64. y

4 62

2

4

x

−2


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


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.


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.


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


Chapter 6


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).


Chapter 6



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.


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


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


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


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


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.


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


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


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.


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

.


Chapter 6


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.


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).


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.


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.


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


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


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.


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


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


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.



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.


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.


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.


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


100 followed by a translation 1 unit

down of the graph of f. The graph of g could be a horizontal


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


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 .


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


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


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.


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



8

−10

−6

8




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


10

−8

−10

12




Chapter 6


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



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.


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.


Chapter 6


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. ✗


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.


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)


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.


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


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


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.


solution is x = 4.


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.


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.


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


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.


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


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


d. A = Pert

1000 = 100e0.06t

10 = e0.06t

ln 10 = ln e0.06t

ln 10 = 0.06t

ln 10 —

0.06 = t


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


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 .


Chapter 6

54. −4 log5 x − 5 ≥ 3

−4 log5 x ≥ 8

log5 x ≤−2

5 log5 x ≤ 5−2

x ≤ 1 — 25


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


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.


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.


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.


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.


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.


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.


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).


y = abx.

2 = ab1

32 = ab3

Step 2 Solve for a in Equation 1 to obtain a = 2 — b and


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).

⤻ ⤻ ⤻


Chapter 6


y = abx.

16 = ab2

2 = ab5


b2 and


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


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.



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.



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.

⤻ ⤻⤻ ⤻⤻


Chapter 6


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.


y = abx.

3 = ab1

12 = ab2



12 = ( 3 — b ) b2

12 = 3b


b = 3 —

4 .

So, the exponential function is y = 3 — 4 (4)x.


y = abx.

24 = ab2

144 = ab3


b2 and

substitute this expression for a into Equation 2.

144 = ( 24 —

b2 ) b3

144 = 24b


— b = 24

— 36

= 2 — 3 .


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


4 = ( 1 — b3 ) b5

4 = b2


b3 = 1 — 23 = 1 —

8 .



y = abx.

27 = ab3

243 = ab5


b3 and


243 = ( 27 —

b3 ) b5

243 = 27b2

9 = b2


— b3 = 27

— 33 = 1.

So, the exponential function is y = 1 ⋅ (3x) = 3x.

⤻⤻ ⤻⤻ ⤻⤻


Chapter 6


y = abx.

2 = ab1

50 = ab3



50 = ( 2 — b ) b3

50 = 2b2

25 = b2


b = 2 —

5 .



y = abx.

40 = ab1

640 = ab3


b and


640 = ( 40 —

b ) b3

640 = 40b2

16 = b2


— 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


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


b2 and


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


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


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.


Chapter 6

19. Sample answer:


x

y

20

40

60

80

4 62



(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 .


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:


x

y

200

400

600

800

4 62



(1, 22) and (2, 39). Substitute the coordinates of

these points into y = ab x.

22 = ab1

39 = ab2


b .


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


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).


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


function that models the data is y = 14.29(1.68x).


Chapter 6


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.


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


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


y = 2.94(1.07)x.


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.


— 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


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.


— 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


y = 9.03(1.99)x.


x 1 4 7 10 13

ln y 1.19 2.31 3.42 4.53 5.64


ln y

4 6 8 10 1220

6

2

0

4

x



such as (1, 1.19) and (4, 2.31). Use these points to write an

equation of the line. Then solve for y.


— 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.


y = 2.27(1.44) x.


x −13 −6 1 8 15

ln y 2.28 2.50 2.72 2.94 3.17


8 124−4−8

4

2

x

ln y

−12



such as (−13, 2.28) and (1, 2.72). Use these points to write

an equation of the line. Then solve for y.


— 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


y = 14.73(1.03)x.


Chapter 6


x −8 −5 −2 1 4

ln y 0.33 0.51 1.67 1.86 2.08


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.


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.


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.


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.


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


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.


5. e4 ⋅ e11 = e4+11

= e15


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.



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


Chapter 6


x

are inverse



x

in the line y = x.


y = ( 1 — 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


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


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.


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


34. 6x > 12

log6 6x > log6 12

x > log6 12

The solution is x > log6 12. Because log6 12 ≈ 1.39, the


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


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.


— 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


y = 3.60(1.43)x.


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.


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.


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


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


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°.

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