Exploring Exponential Models

ALGEBRA 2 LESSON 8-1ALGEBRA 2 LESSON 8-1

Evaluate each expression for the given value of x.

1. 2x for x = 3 2. 4x+1 for x = 1

3. 23x+4 for x = –1 4. 3x3x–2 for x = 2

5. for x = 0 6. 2x for x = –212

(For help, go to Lesson 1-2.)

Exploring Exponential ModelsExploring Exponential Models

8-1

x

1. 2x for x = 3: 23 = 2 • 2 • 2 = 8

2. 4x+1 for x = 1: 41+1 = 42 = 4 • 4 = 16

3. 43x+4 for x = –1: 23(–1)+4 = 2–3+4 = 21 = 2

4. 3x3x–2 for x = 2: 32 • 32 – 2 = 32 • 30 = 3 • 3 • 1 = 9

5. for x = 0: = 1

6. 2x for x = –2: 2–2 = = =

12

1 22

1 2 • 2

14

Solutions



8-1

x 12

0

Graph y = 3x.

Step 2: Graph the coordinates. Connect the points with a smooth curve.



8-1

x 3x y

–3 3–3 = .037

–2 3–2 = .1

–1 3–1 = .3

0 30 1

1 31 3

2 32 9

3 33 27

1 271913

Step 1: Make a table of values.

The population of the United States in 1994 was almost 260

million with an average annual rate of increase of about 0.7%.

a. Find the growth factor for that year.

b = 1 + r = 1 + 0.007 Substitute 0.7%, or 0.007 for r. = 1.007 Simplify.

Relate: The population increases exponentially, so y = abx

Define: Let x = number of years after 1994.Let y = the population (in millions).

b. Suppose the rate of growth had continued to be 0.7%. Write a function to model this population growth.



8-1

Write: y = a(1.007)x

260 = a(1.007)0 To find a, substitute the 1994 values: y = 260, x = 0.

260 = a • 1 Any number to the zero power equals 1.

260 = a Simplify.

y = 260(1.007)x Substitute a and b into y = abx.

y = 260(1.007)x models U.S. population growth.



(continued)

8-1

Write an exponential function y = abx for a graph that

includes (1, 6) and (0, 2).

y = abx Use the general term.

6 = ab1 Substitute for x and y using (1, 6).

= a Solve for a.6b

2 = b0 Substitute for x and y using (0, 2) and for a using .6b

6b

2 = • 1 Any number to the zero power equals 1.

2 = Simplify.

b = 3 Solve for b.

6b6b



8-1

The exponential for a graph that includes (1, 6) and (0, 2) is y = 2 • 3x.

a = Substitute 3 for b.

a = 2 Simplify.

y = 2 • 3x Substitute 2 for a and 3 for b in y = abx.

63



(continued)

8-1

a = Use your equation for a.6b

Without graphing, determine whether the function y = 3

represents exponential growth or decay.

23

Since b < 1, the function represents exponential decay.



8-1

In y = , b = .3 23

23

x

x

Step 1: Make a table of values.

x –3 –2 –1 0 1 2 3

y 288 144 72 36 18 9 412

Graph y = 36(0.5)x. Identify the horizontal asymptote.

Step 2: Graph the coordinates. Connect the points with a smooth curve.

As x increases, y approaches 0.

The horizontal asymptote is the x-axis, y = 0.



8-1

Suppose you want to buy a used car that costs $11,800. The

expected depreciation of the car is 20% per year. Estimate the

depreciated value of the car after 6 years.

The decay factor b = 1 + r, where r is the annual rate of change.

b = 1 + r Use r to find b.

= 1 + (–0.20) = 0.80 Simplify.

Write an equation, and then evaluate it for x = 6.

Define: Let x = number of years. Let y = value of the car.

Relate: The value of the car decreases exponentially; b = 0.8.



8-1

Write: y = ab x

11,800 = a(0.8)0 Substitute using (0, 11,800).

11,800 = a Solve for a.

The car’s depreciated value after 6 years will be about $3,090.

y = 11,800(0.8)6 Evaluate for x = 6.

3090 Simplify.

y = 11,800(0.8)x Substitute a and b into y = abx.



(continued)

8-1



8-1

1.

2.

3.

4.

6.

7.

5.8.

9. a. 1.0126b. y = 6.08(1.0126)x,

where x = 0 corresponds to 2000

pages 426–429 Exercises



8-1

10. y = 0.5(2)x

11. y = 2.5(7)x

12. y = 8(1.5)x

13. y = 5(0.6)x

14. y = 3(0.5)x

15. y = 24

16. exponential growth

17. exponential decay







13

24–31. y = 0 is the horizontal asymptote.

24.

25.

26.

27.

28.

29.

x

44. y = 30,000(0.7)x for car 1; y =15,000(0.8)x for car 2; car 2 will be worth more.

45. a. y = 80(0.965)x b. 56 animalsc.

about 47 years

46. 1.70

47. 6

48. 0.25

49. 0.45

30.

31.

32. y = 100(0.5)x; 1.5625

33. y = 12,000(0.9)x; 6377

34. y = 12,000(0.1)x; 0.012

35. a. y = 6500(0.857)x

b. about $4091.25

36. a. Tokyo: 26,444,000,Mexico City:18,850,649,Bombay: 23,148,579,São Paulo:

19,496,367b. yes; Tokyo, Bombay,

São Paulo, Mexico City

37. 63% increase

38. 30% increase

39. 35% decrease

40. 70% increase

41. 87.5% decrease

42. 75% decrease

43. a. 5.6%b. 0.0017%



8-1



8-1

60. C

61. H

62. B

63. [2]

[1] general form of graph correct,

but not exact

59. a. A negative growth rate would be represented by adding the negative rate to 1.

b. Armenia: y = 9.2(1.06)x,

Canada: y = 688.3(1.03)x,

Oman: y = 18.6(.915)x,

Paraguay: y = 19.8(.995)x,

x in each equation represents the number of years since 1998.

c. Armenia: $13.8 billion,

Canada: $846.5 billion,

Oman: $10.0 billion,

Paraguay: $19.1 billion

50. 1.125

51. 0.999

52. 1.001

53. 2

54. y = 34(1.26)x, where xrepresents the number of years since 1995.

55. Check students’ work.

56. about $42,140

57. C

58. B; the graph shows a decreasing function, so b < 1, which eliminates A. The graph is always positive, which eliminates C.



8-1

67.

68. 6n2 5n

69. 84r 2 9r 2

70.

71. Answers may vary.Sample:y = x3 – 5x2 + 4x

72. Answers may vary.Sample:

y = x3 – 7x – 6

[3] appropriate methods with one computational error[2] starts problem

correctly, but fails to finish it properly[1] correct answer,

without work shown

65.

66.

64. [4] y = abx

54 = ab2

a =

2 = b

2 =

2b = 54

b = 27

b = 27b = 9

a =

a =

y = 9x

54b2

54b2

1

2

54

b 3

2

23

2 x 3x

33

23

22

3

5492

23



8-1

73. Answers may vary.Sample:

y = x3 – 7x2 + 10x

74. y = x2 + 2

75. y = –5x2 + 2

76. y = 2x2 + 2

77. y = 10x2 + 2

78. y = x2 + 2

79. y = 3x2 + 2

80. y = – x2 + 2

81. y = –x2 + 2

32

12

Sketch the graph of each function. Identify the horizontal asymptote.

1. y = (0.8)x 2. y =

Without graphing, determine whether each equation represents exponential growth or decay.

3. y = 15(7)x 4. y = 1285(0.5)x

Write an exponential function for a graph that includes thegiven points.

5. (0, 0.5), (1, 3) 6. (–1, 5), (0.5, 40)

14

growth decay

y = 0.5(6)x y = 20(4)x

y = 0 y = 0



8-1

x

(For help, go to Lesson 2-6, 5-3, and 7-4.)


Properties of Exponential FunctionsProperties of Exponential Functions

8-2

Write an equation for each translation.

1. y = | x |, 1 unit up, 2 units left

2. y = –| x |, 2 units down

3. y = x2, 2 units down, 1 unit right

4. y = –x2, 3 units up, 1 unit left

Write each equation in simplest form. Assume that all variablesare positive.

5. y = 6. y = 7. y =

8. Use the formula for simple interest I = Prt. Find the interest for a principal of $550 at a rate of 3% for 2 years.

54

–x

17

–x56x4 –7 6



8-2

1. y = | x |, 1 unit up and 2 units left: y = | x + 2 | + 1

2. y = –| x |, 2 units down: y = –| x | – 2

3. y = x2, 2 units down and 1 unit right: y = (x – 1)2 – 2

4. y = –x2, 3 units up and 1 unit left: y = –(x + 1)2 + 3

5. y = x ; y = x (4); y = x–5 or y =

6. y = x ; y = x (–7); y = x1; y = x

7. y = x ; y = x (6); y = x5

8. I = Prt = $550(0.03)(2) = $16.50(2) = $33

54

– 1 x5

17

–

56

Solutions

4

–7

6

54

–

17

–

56

Graph y = 3 • 2x and y = –3 • 2x. Label the asymptote of

each graph.

x y = 3 · 2x y = –3 · 2x

–3

–2

–1

0 3 –3

1 6 –6

2 12 –12

3 24 –24

3834

121

3834

121

–

–

–

Step 1: Make a table of values Step 2: Graph each function.



8-2

Step 1: Graph y = 6 . The horizontal asymptote is y = 0.12

x

Graph y = 6 and y = 6 – 2.

So shift the graph of the present function 3 units right and 2 units down.

The horizontal asymptote is y = –2.



8-2

x x–3

Step 2: For y = 6 – 2, h = 3 and k = –2.12

x–3

12

12

16

Define: Let y = the amount of technetium-99m.Let x = the number of hours elapsed.

Then x = the number of half-lives.

Since a 100-mg supply of technetium-99m has a half-life of 6 hours, find the amount of technetium-99m that remains from a 50-mg supply after 25 hours.

Relate: The amount of technetium-99m is an exponential function of the number of half-lives. The initial amount is 50 mg. The decay

factor is . One half-life equals 6 h.12



8-2

After 25 hours, about 2.784 mg of technetium-99m remains.



(continued)

8-2

= 50 Simplify.

2.784

12

4.16

Write: y = 50

y = 50 Substitute 25 for x.

16

• 25

12

x

16

12

Graph y = ex. Evaluate e3 to four decimal places.

Step 1: Graph y = ex.

The value of e3 is about 20.0855.



8-2

Step 2: Find y when x = 3.

Suppose you invest $100 at an annual interest rate of 4.8%

compounded continuously. How much will you have in the account

after 3 years?

A = Pert

= 100 • e0.048(3) Substitute 100 for P, 0.048 for r, and 3 for t.

= 100 • e0.144 Simplify.

= 100 • (1.154884) Evaluate e0.144.

You will have about $115.49 in the account after three years.

115.49 Simplify.



8-2


1–8. Asymptote is y = 0.

1.

2.

3.

4.

5.

6.

7.

8.



8-2



8-2

9.

10.

11.

12.

13.

14.

15. y = 50 ; 0.85 mg

16. y = 200 ; 0.43 mg

17. y = 24 ; 0.64 mg

18. 20.0855

19. 403.4288

20. 0.1353

21. 1

22. 12.1825

23. 15.1543

24. $2330.65

25. $448.30

26. $1819.76

27. 0

12

1

14.3

12

1

8.14

12

1

5730

x

x

x



8-2

28. 1

29. If c < 0, the graph models exponential decay. If c = 0, the graph is a horizontal line. If c > 0, the graph models exponential growth.

30. $6168.41

31. a. Answers may vary. Sample: y = –

2(1.3)x

b. Answers may vary. Sample: I am in

debt for $2 and my debt

is growing at a rate

of 30% per year.

c. The graph of exponential decay approaches the asymptote y = 0 as x increases. The graph of negative exponential growth approaches the asymptote y = 0 as x decreases.

32.y = 4 ;

y = 4 + 3

33.y = –3x;

y = –3x–8 + 2

1212

34. y = (2)x;

y = (2)x–6 – 7

35. y = –3 ;

y = –3 – 1

36. 75.0 pascals37. 8.7 yr38. A deficit that is

growing exponentially is modeled by y = abcx, where a < 0, and either b > 1 and c > 0 or 0 < b < 1 and c < 0.

1212

13

13

x

x + 4

x + 5

x

39. a. GDP = 8.511(1.038)t where t = 0 corresponds to 1998 and 8.511 is trillions.b. It almost doubles.

(about 195.7%)c. In 9 years, the growth is about 40%.

40. a. $2501.50

b. $3.15 more

41. $399.97







48. a. y = 8001 – 3x, where y is the number of uninfected people and x represents

days.

b. 5814 people

c. about 9 days

49. a. about 10 names; about 24

names

b. Graphically, it will never happen; the graph has y = 30 as an asymptote. (In reality, you would be close to knowing all the names in about 21 days.)

c. Answers may vary. Sample: I learn names pretty quickly; my learning rate might be 0.4.



8-2

58. y = –5(2)x

59. y = 8(0.5)x

60. y = 70(10)x

61. y = 10,000(0.8)x

62. 6 5

63. –

64. 5( )

65. 2( )

66. 3

67. 11 7

68. 15 – 6 6

69. 3 + 5

70. x – 3, R –1

34

3 + 5

2 + 84 4

71. x2 – x – 6

72. x2 – 1

73. x2 + x + 1

74. 13x + 1

75. 3x2 – 7x + 2

76. x – y = 5, x + z = 5,–y + z = 5

77. x + y = –2, x + 4z = –2, y + 4z = –2

78. x + y = 8, x – 2z = 8, y – 2z = 8



8-2

50. a. 2928 m3

b. V = 2928 – 15(2x –

1)c. ninth weekend

51. A

52. I

53. C

54. F55. [2] A = Pert

8000 = Pe(0.06)(4)

= P

P = $6293.02[1] answer only, without

work shown

56. y = 3x

57. y = –2(4)x

8000 e(0.06)(4)

79. x – y = 8, x + 2z = 8,–y + 2z = 8

80. 3x + y = –18,x + 3z = –6,y + 9z = –18

81. –2x + y = 10, –2x – 5z = 10, y – 5z = 10



8-2

Describe how the graph of each function relates to its parent function. Then graph the function.

1. y = –5x – 1 2. y = 3(5)x – 2 + 1

reflected over the x-axis and translated down 1

rises more steeply, is translated 2 to the right and up 1

3. Use the formula A = Pert to find the amount in a continuously compounded account where the principal is $2000 at an annual interest rate of 5% for 3 years.

4. Use the graph of y = ex to evaluate e to four decimal places.1

3

about $2324

1.3956



8-2

(For help, go to Lessons 7-1 and 7-7.)


Logarithmic Functions as InversesLogarithmic Functions as Inverses

8-3

Solve each equation.

1. 8 = x3 2. x = 2 3. 27 = 3x 4. 46 = 43x

Graph each relation and its inverse on a coordinate plane.

5. y = 5x 6. y = 2x2 7. y = –x3 8. y = x

14

12



8-3

2. x = 2; since 16 = = 2, x = 16

6. y = 2x2

Inverse:x = 2y2

y2 =

y2 = ±

8. y = xInverse:

x = y

y = 2x

x2

1. 8 = x3; since 23 = 8, x = 2

3. 27 = 3x; since 33 = 27, x = 3

5. y = 5xInverse:x = 5y

y = x

7. y = –x3

Inverse:x = –y3

y3 = –xy =

15

3–x

1

4

1

4

416

12

12

4. 46 = 43x; since 6 = 3(2), x = 2

Solutions

x2

Compare the amount of energy released in an earthquake

that registers 6 on the Richter scale with one that registers 3.

= 306–3 Division Property of Exponents

= 303 Simplify.

= 27,000 Use a calculator.

The first earthquake released about 27,000 times as much energy as the second.

Write a ratio.E • 306

E • 303

= Simplify.306

303



8-3

Write: 32 = 25 in logarithmic form.

If y = bx, then logb y = x. Write the definition.

If 32 = 25, then log2 32 = 5. Substitute.

The logarithmic form of 32 = 25 is log2 32 = 5.



8-3

Evaluate log3 81.

Let log3 81 = x.

Log3 81 = x Write in logarithmic form.

81 = 3x Convert to exponential form.

34 = 3x Write each side using base 3.

4 = x Set the exponents equal to each other.

So log3 81 = 4.



8-3

The pH of an apple is about 3.3 and that of a banana is about

5.2. Recall that the pH of a substance equals –log[H+], where [H+] is the

concentration of hydrogen ions in each fruit. Which is more acidic?

The [H+] of the apple is about

5.0 10– 4.

The [H+] of the banana is

about 6.3 10– 6.

The apple has a higher concentration of hydrogen ions, so it is more acidic.



8-3

Apple

pH = –log[H+]

3.3 = –log[H+]

log[H+] = –3.3

[H+] = 10–3.3

5.0 10– 4

[H+] = 10–5.2

pH = –log[H+]

5.2 = –log[H+]

log[H+] = –5.2

Banana

6.3 10– 6

Graph y = log4 x.

By definition of logarithm, y = log4 x is the inverse of y = 4x.

Step 1: Graph y = 4x.

Step 2: Draw y = x.

Step 3: Choose a few points on 4x. Reverse the coordinates and plot the points of y = log4 x.



8-3

Step 2: Graph the function by shifting the points from the table to the right 1 unit and up 2 units.

Graph y = log5 (x – 1) + 2.

Step 1: Make a table of values for the parent function.

11251

2515

1

5

–3

–2

–1

0

1

x y = log5 x



8-3



8-3


1. The earthquake in Missouri released about 1.97 timesmore energy.

2. The earthquake in Chile released about 231 times more energy.

3. The earthquake in Missouri released about 8,759,310 times more energy.

4. The earthquake in Missouri released about 30 times more energy.

5. The earthquake in Alaska released about 83 times more energy.

6. log7 49 = 2

7. 3 = log 1000

8. log5 625 = 4

9. log = –1

10.2 = log8 64

11. log 4 = –2

12.3 = log ( )

1

10

1

2

1

3

1 27

13. –2 = log 0.01

14. 4

15.

16. 1

17.

18. 3

19.

20. undefined

21. 2

22. 5

23. 1

24. 4

25. 3

12

32

12

26. 6.3 10–

6

27. 6.3 10–

3

28. 1.0 10–

8

29. 7.9 10–

4

30. 5.0 10–

7

31. 1.3 10–

5

32. 1.0 10–

4

33. 4.0 10–

6

34. 2.5 10–

4

35.

36.

37.

38.

39.

40.

41. 0.6990; 0

42. –4.2147; –5

43. –1.0969; –2

44. 2.3010; 2

45. –0.7782; –1



8-3



8-3

46. 1.2435; 1

47. 7.1139; 7

48. 0.5119; 0

49. apple juice: 3.5, acidic; buttermilk: 4.6, acidic; cream: 6.6, acidic; ketchup: 3.9, acidic; shrimp sauce: 7.1, basic; strained peas: 6, acidic

50. The error is in the second line. It should read 3 = 27x; the correct answer is .1

3

51. First rewrite y = log1 x as 1y = x. For any real number y, x = 1.

52. Answers may vary. Sample: y = log3 x; y = 3x

53. 128 = 27

54. 0.0001 = 10–4

55. 16,807 = 75

56. 6 = 61

57. 1 = 40

58. = 3–2

59. = 2–1

60. 10 = 101

61. 8192 = 213

1912



8-3

62. a. buffalo bone: 9826 to 10,128 years old, bone fragment: 9776 to

10,180 years old, pottery shard: 0

to 183 years old, charcoal:

9718 to 10,242 years old, spear shaft: 9776 to 10,263 years oldb. The pottery shard;

answers may vary. Samples: the pottery may be from a later civilization, or the mass or the beta radiation emissions may have

been measured incorrectly.

63. y = 4x

64. y = 0.5x

65. y = 10x

66. y = 2x–1

67. y = 10x – 1

68. y = 10x–1

69. y = 10x + 2

70. y = 5

71. y = 3

72.

x

2

1

x

73.

74.

75. domain {x|x > 0}, range: all reals

76. domain {x|x > 0}, range: all reals

77. domain {x|x > 3},range: all reals



80. domain {x|x > –1},range: all reals



83. domain {x|x > t },range: all reals

84. 100

85. 70

86. 60

87. 20

88. 10

89. a. III

b. I

c. IV

d. II

90. a. II

b. III

c. I

91. D

92. I

93. C

94. A

95. B

96. C

97.

y = –100



8-3

98.

y = 0

99.

y = 9



8-3

w3

3t 2

5z8

nx p

12

12

100.

101. 4

102.

103.

104. –2, –4

105. –0.162, 6.162

106. –1.217, 0, 8.217

107. (2x – 3)(2x – 1)

108. ( b – 2)( b + 2)

109. (5x – 2)(x + 3)

Evaluate each logarithm.

1. log232 2. log10(–100) 3. log5

Write each equation in exponential form.

4. log 0.01 = –2 5. log327 = 3

6. Graph y = log6(x – 3) + 1.

5 undefined –2

10–2 = 0.01 33 = 27



8-3

1 25


Simplify each expression.

1. log24 + log28 2. log39 – log327 3. log216 ÷ log264

Evaluate each expression for x = 3.

4. x3 – x 5. x5 • x2 6. 7. x3 + x2x6

x9


Properties of LogarithmsProperties of Logarithms

8-4



8-4

1. log24 = x log28 = y

2x = 4 2y = 8 x = 2 y = 3

log24 + log28 = 2 + 3 = 5

2. log39 = x log327 = y

3x = 9 3y = 27 x = 2 y = 3

log39 – log327 = 2 – 3 = –1

3. log216 = x log264 = y

2x = 16 2y = 64 x = 4 y = 6

log216 ÷ log264 = 4 ÷ 6 = = 46

23

4. x3 – x for x = 3: 33 – 3 = 27 – 3 = 24

5. x5 • x2 for x = 3: 35 • 32 = 3(5+2) = 37 = 2187

6. for x = 3: = 3(6–9) = 3–3 =

=

7. x3 + x2 for x = 3: 33 + 32 = 27 + 9 = 36

x6

x9

36

99

1 33

1 27

Solutions

State the property or properties used to rewrite each

expression.

a. log 6 = log 2 + log 3

Product Property: log 6 = log (2•3) = log 2 + log 3

b. logb = 2 logb x – logb yx2

y

Quotient Property: logb = logb x2 – logb yx2

y

Power Property: logb x2 – logb y = 2 logb x – logb y



8-4

Write each logarithmic expression as a single logarithm.

a. log4 64 – log4 16

= log4 4 or 1 Simplify.

log4 64 – log4 16 = log4 Quotient Property6416

b. 6 log5 x + log5 y

6 log5 x + log5 y = log5 x6 + log5 y Power Property

= log5 (x6y) Product Property

So log4 64 – log4 16 = log4 4, and 6 log5 x + log2 y = log5 (x6y).



8-4

Expand each logarithm.

a. log7 tu

b. log(4p3)

log(4p3) = log 4 + log p3 Product Property

= log 4 + 3 log p Power Property



8-4

log7 = log7 t – log7 u Quotient Propertytu

Manufacturers of a vacuum cleaner want to reduce its sound

intensity to 40% of the original intensity. By how many decibels would

the loudness be reduced?

Relate: The reduced intensity is 40% of the present intensity.

Define: Let l1 = present intensity. Let l2 = reduced intensity.Let L1 = present loudness. Let L2 = reduced loudness.

Write: l2 = 0.04 l1

L1 = 10 log

L2 = 10 log

l1l0l2l0



8-4

L1 – L2 = 10 logl1l0

l2l0

– 10 log Find the decrease in loudness L1 – L2.

= 10 logl1l0

0.40l1l0

– 10 log Substitute l2 = 0.40l1.

= 10 logl1l0

– 10 log 0.40 • l1l0

Product Property= 10 logl1l0

– 10 ( log 0.40 + log ) l1l0

= 10 logl1l0

– 10 log 0.40 – 10 logl1l0

Distributive Property

= –10 log 0.40 Combine like terms.

4.0 Use a calculator.

The decrease in loudness would be about 4 decibels.



(continued)

8-4



8-4


1. Product Property

2. Quotient Property

3. Power Property

4. Power Property

5. Power Property,

Quotient Property

6. Power Property

7. Power Property,

Quotient Property

8. Power Property,

Product Property

9. Power Property,

Quotient Property

10. Power Property, Product Property

11. log 14

12. log2 3

13. log 972

14. log

15. log

16. log

17. log6 5x

18. log7

19. 3 log x + 5 log y

23m4

n 5 2k

xyz

20. log7 22 + log7 x + log7 y + log7 z

21. log4 5 + log4 x

22. log 3 + 4 log m – 2 log n

23. log5 r – log5 s

24. 2 log3 2 + 2 log3 x

25. log3 7 + 2 log (2x – 3)

26. 2 log a + 3 log b –

4 log c

27. log 2 + log x – log y

28. 1 + log8 3 + log8 a

29. log s + log 7 – 2 log t

12

12

12

12

12

52

12



8-4

t s

30. –logb x

31. 9 dB

32. 13 dB

33. –2

34. 1

35. 6

36. 2

37. 2

38. 1

39. 1

40. –2

41. 1

12 t s

12

12

51. –1.398

52. 1.398

53. –0.7782

54. 1.5564

55. 0.3495

56. 12 dB

57. 0.00001

58. True; log2 4 = 2 and

log2 8 = 3.

59. False; log3 3 =

log3 3 , not log3 .

121

2

12

12

32

42. The coefficient is missing in log4 s;

log 4 = log4 = (log4 t – log4 s)

= log4 t – log4 s.

43. Answers may vary. Sample: log 150 = log 15 + log 10

44. 1.3803

45. 1.4772

46. 1.2042

47. 2.097

48. 0.1761

49. –0.0969

50. –0.6021



8-4

60. True; it is an example of the Power Property since 8 = 23.

61. False; the two logs have different bases.

62. False; this is not an example of the Quotient Property. log (x – 2) log x – log 2.

63. False; logb = logb x – logb y.

64. False; the exponent on the left means the log x, quantity squared, not the log of x2.

xy

65. False; log4 7 – log4 3 = log4 , not log4 4.

66. True; log x + log (x2 + 2) = log x(x2 + 2), which equals log (x3 + 2x).

67. False; the three logs have different bases.

68. True; the power and quotient properties are used correctly.

69. True; the left side equals

logb ( • 43), which

equals logb 8.

70. 102 dB

73

18

71. No; the expression (2x + 1) is a sum, so it is not covered by the Product, Quotient, or Power properties.

72. The log of a product is equal to the sum of the logs. log (MN) = log M + log N. So log (5 • 2) = log 10 = 1, log 5 • log 2 (0.7)(0.3) = 0.21, which is not equal to 1.

=/



8-4

81. log 2 + log 4 +

log r – log s

82. logb x + logb y –

logb z

83. log4 x + log4 y –

log4 z – 4 log4 w

84. log (x2 – 4) –

2 log (x + 3)

85. log x + log 2 – log y

12

12

12

23

25

52

72

12

12

14

86. log3 x + log3 y – 6 log3 z

87. log7 (r + 9) –

2 log7 s – log7 t

88. v = logb Nbv = NMN = bu • bv = bu+v

logb MN = u + vlogb MN = logb M + logb N

12

13

42x

2 yz3

272mxn

p

x2 y3

z5

3 4

4

4

32

1

y

z

3 x5

73. log3

74. logx

75. log

76. log4

77. logb

78. log

79. 3 log 2 + log x – 3 log 5

80. 3 log m – 4 log n + 2 log p



8-4

89. 1. u = logb M (given)2. bu = M (Rewrite in

exponential form.)3. (bu)x = Mx (Raise

each side to x power.)4. bux = Mx (Power

Property of exponents)5. logb bux = logb Mx

(Take the log of each side.)6. ux = logb Mx (Simplify.)7. logb Mx = x • logb M

(substitution)

90. 1. u = logb M (given)2. bu = M (Rewrite in

exponential form.)3. v = logb N (given)4. bv = N (Rewrite in

exponential form.)

5. = = bu–v

(Quotient Property of Exponents)

6. logb = logb bu–v (Take the log of

each side.)

7. logb = u – v (Simplify.)

8. logb = logb M –

logb N (substitution)

MN

bu

bv

MN

MN

MN

91. B

92. G

93. D

94. [2] By the Quotient

Property, log5

= log5

1.4307 –

1.8614 = –0.4307.

[1] correct answer, without work

shown

12

1020



8-4

95. [4] 1) log = log 24 –

log 2; Quotient

Property2) log 2 • 6 = log 2 +

log 6; Product Property

3) log 144 =

log 144; Power

Property4) log 3 • 22 = log 3

+ 2 log 2; Product

and Power Properties

[3] log 12 written three ways

242

1

212

[2] log 12 written two ways OR

written 4 ways but

properties not named

[1] log 12 written only 2 ways and

properties not named

96. log7 49 = 2

97. 3 = log5 125

98. log8 = –

99. –3 = log5

100. 8, –8

14

23 1 125

101.

102. 2

103. –

104. i, –4i

105. –2i, –4 – i

106. – 2,– i – 1

107.

108. 2i + 3, –i

647

3, 5

7, 11

Write each expression as a single logarithm. State the property you used.

1. log 12 – log 3 2. 3 log115 + log117

Expand each logarithm.

3. logc 4. log3x4

Use the properties of logarithms to evaluate each expression.

5. log 0.001 + log 100 6. logyy

log 4; Quotient Property log11(53 • 7); Power Property and Product Property

logca – logcb 4 log3x

–1 12



8-4

12

ab

Evaluate each logarithm.

1. log981 • log93 2. log10 • log39

3. log216 ÷ log28 4. Simplify 125 .



Exponential and Logarithmic EquationsExponential and Logarithmic Equations

8-5

2

3–



8-5

4. 125 = = = = = 1

3

1

(125 ) 2

1. log981 = x log93 = y 9x = 81 9y = 3 x = 2 y =

log981 • log93 = 2 • = 1

3. log216 = x log28 = y 2x = 16 2y = 8 x = 4 y = 3

log216 ÷ log28 = 4 ÷ 3 = or 1

1212

43

13

2. log10 = x log39 = y 10x = 10 3y = 9 x = 1 y = 2

log10 • log39 = 1 • 2 = 2

2

3

1

125

1 52

1 253

1252

1

( )

Solutions

2

3–

Solve log 52x = 16.

52x = 16

52x = log 16 Take the common logarithm of each side.

2x log 5 = log 16 Use the power property of logarithms.

x = Divide each side by 2 log 5.log 162 log 5


Check: 52x 16

52(0.8614) 16



8-5

Use the Change of Base Formula to evaluate log6 12. Then

convert log6 12 to a logarithm in base 3.

log6 12 = Use the Change of Base Formula.log 12log 6

1.387 Use a calculator.1.07920.7782

log6 12 = log3 x Write an equation.

1.387 log3x Substitute log6 12 = 1.3868

1.387 Use the Change of Base Formula.log xlog 3



8-5


1.387 • log 3 log x Multiply each side by log 3.

1.387 • 0.4771 log x Use a calculator.

0.6617 log x Simplify.

x 100.6617 Write in exponential form.

The expression log6 12 is approximately equal to 1.3869, or log3 4.589.



(continued)

8-5

Solve 52x = 120.

52x = 120

log5 52x = log5 120 Take the base-5 logarithm of each side.

2x = log5 120 Simplify.

2x = Use the Change of Base Formula.log 120

log 5

x 1.487 Use a calculator to solve for x.



8-5

Solve 43x = 1100 by graphing.

The solution is x 1.684

Graph the equations y = 43x and y = 1100.

Find the point of intersection.



8-5

The population of trout in a certain stretch of the Platte River

is shown for five consecutive years in the table, where 0 represents the

year 1997. If the decay rate remains constant, in the beginning of

which year might at most 100 trout remain in this stretch of river?

Time t 0 1 2 3 4

Pop. P(t) 5000 4000 3201 2561 2049

Step 1: Enter the data into your calculator.



8-5

Step 2: Use the Exp Reg feature to find the exponential function that fits the data.

Step 3: Graph the function and the line y = 100.

Step 4: Find the point of intersection.

The solution is x 18, so there may be only 100 trout remaining in the beginning of the year 2015.



(continued)

8-5

Solve log (2x – 2) = 4.

log (2x – 2) = 4

2x – 2 = 104 Write in exponential form.

2x – 2 = 10000

x = 5001 Solve for x.

log 104 = 4

log 10,000 4

log (2 • 5001 – 2) 4

Check: log (2x – 2) 4



8-5

Solve 3 log x – log 2 = 5.

3 log x – log 2 = 5

x3

2Log ( ) = 5 Write as a single logarithm.

x3

2 = 105 Write in exponential form.

x3 = 2(100,000) Multiply each side by 2.

x = 10 200, or about 58.48.3

The solution is 10 200, or about 58.48.3



8-5

19. 2.3219

20. 0.8496

21. 0.0499

22. 3.0101

23. 1.0219

24. 0.9746

25. 0.2009

26. 5.2379


1. 1.5850

2. 2.1240

3. 2.7320

4. 3.0101

5. 3

6. 3.4650

7. 0.9534

8. 0.3579

9. 3.2056

10. 0.2720

11. 3.1699; log8 729

12. 1.5; log8 22.627

13. 3.6309; log8 1901.3

14. 2.5643; log8 206.93

15. 3.1827; log8 748.56

16. 2.8074; log8 343

17. 3.8737; log8 3149.6

18. 0.0792; log8 1.1790



8-5

43. 100,000 5, or 223,606.8

44. 5

45.

46. 1357.2

47. 7

27. 0.5690

28. 1.2871

29. 4.7027

30. 14.4894

31. about 7.1 years

32. about 2018

33. 0.05

34. , or 0.3162 10

10

160 1

4



8-5

35. 33

36. 10,000

37. , or 0.0167

38. 12

39. 10, or 3.1623

40. 100 10 – 1, or 315.2

41. 2

42. 3 108



8-5

59. 3

60. –

61. a. Florida growth factor = 1.0213, y =

15,982,378 • (1.0213)x;

New York growth factor =

1.0054, y = 18,976,457 • (1.0054)x

b. 2011

48. a. 18.9658

b. 18.9658

c. Answers may vary. Sample: You don’t have to use the Change of Base formula with the base-10 method, but there is less algebra with the base-2 method.

49. 5.1

50. 20 = 8(1.2)x, 5 years

51. 2 = 10( ) , 2.7 min

52. 75 = 125(0.88)x, 4 years

53. –1

54. 3

55.

56. 3

57.

58. –2

12

x

1.17

12

13

12



8-5

65. Answers may vary. Sample: log x = 1.6 101.6 = x, x 39.81

66. Answers may vary. Sample: A possible model is y = 1465(1.0838)x where x = 0 represents 1991; the growth is probably exponential and 1465(1.0838)10 3276; using this model, there will be 10,000 manatees in about 2015.

62. a. Texas growth factor = 1.0208, y =

20,851,820 (1.0208)x;

California growth factor =

1.013, y = 33,871,648 • (1.013)x

b. 2063

63. Since Florida’s growth rate is larger than Texas’s growth rate, in theory, given constant conditions, Florida would exceed Texas in about 543 years. However, since no state has unlimited capacity for growth, it is unrealistic to predict over a long period of time.

64. 2 – 1log102

log101 =/



8-5

76. a. 100 (or 1) W/m2, 104 W/m2

b. 10,000 times more intense

77. a. top up: 10–5 W/m2, top down:

10–2.5 W/m2

b. 99.68%

78. a. 10–3 W/m2, 106 W/m2

b. 109 times more intense

79. 2.9315

67. a. x =

b. x = logab =

c. Substituting the

result from

part (a) into

the results from part

(b), or vice

versa, yields

logab = .

This justifies the

Change of Base

Formula for c =

10.

log blog a

log blog a

log blog a

68.

69.

70.

71.

72.

73.

74.

75.

log 2log 7

log 8log 3

log 140log 5

log 3.3log 9

log 3xlog 4

log (1 – x)log 6

log 5log x

log (x + 1)log x

97. a. bassoon, guitar, harp, violin,

viola, cellob. bassoon, guitar,

harp, cello, bassc. harp, violind. harp, violin

98. 478,630 times

99. no; solving

0.65 = for x, the

age in years of the

sample, yields an age

of about 3561 yrs.

100. 5

80. 0.2098

81. 0.6225

82. 625

83. 2.3094

84. 10

85. 0.8505

86. 1.5

87. 7.4168

88. 200.8

89. 2.9615

90. 2.7944

91. 1

92. 500

93. 1.0451

94. 114.3

95. 1.3063

96. 3.0417

x (0.5)5430



8-5



8-5

110. 27.7

111. 249.5

112. 103

113. log 2 + 3 log x – 2 log y

114. log3 x – log3 y

115. 3 log2 3 + 3 log2 x

116. log3 7 + 2 log3 (2x – 3)

117. log4 5 + log4 x

118. log2 5 + log2 a – 2 log2 b

119. x2 + 3x – 1

120. x2 – 3x – 1

121. 3x3 – 3x

122. 1, ±i

101. –4, 2

102. –9, 9

103. 1

104. 20,031 m above sea level

105. a. 91 hours or 4 days b. 0.928 mg or 1.061 mg c. Estimate in hours is more accurate; the days have a

larger rounding error.

106. 1.11

107. 7.30

108. 2.19

109. 138.8

12



8-5

123. ±2, ±2i

124. ± 2, ± 3

125. –2, –1, 3

126. 2(x) + 2(x + 12) = 128; 26 ft

127. = 133; 119679 + x6

Use mental math to solve each equation.

1. 2x = 2. log42 = x 3. 106x = 1

4. Solve 52x = 125.

–3 12

0

32



8-5

18


Use your calculator to evaluate each expression to the nearest thousandth.

1. e5 2. 2e3 3. e–2 4. 5. 4.2e

Solve.

6. log3x = 4 7. log164 = x 8. log16x = 4


Natural LogarithmsNatural Logarithms

8-6

1e



8-6

7. log164 = x

16x = 4

x = 12

1. e5 148.413 2. 2e3 40.171

3. e–2 0.135 4. 0.368

5. 4.2e 11.417

1e

6. log3x = 4

34 = x

81 = x

8. log16x = 4

164 = x

65,536 = x

Solutions

Write 2 ln 12 – ln 9 as a single natural logarithm.

2 ln 12 – ln 9 = ln 122 – ln 9 Power Property

= ln Quotient Property122

9

= ln 16 Simplify.



8-6

Find the velocity of a spacecraft whose booster rocket has a

mass ratio 22, an exhaust velocity of 2.3 km/s, and a firing time of 50 s.

Can the spacecraft achieve a stable orbit 300 km above Earth?

Let R = 22, c = 2.3, and t = 50. Find v.

v = –0.0098t + c ln R Use the formula.

= –0.0098(50) + 2.3 ln 22 Substitute.

–0.49 + 2.3(3.091) Use a calculator.

6.62 Simplify.

The velocity is 6.6 km/s is less than the 7.7 km/s needed for a stable orbit. Therefore, the spacecraft cannot achieve a stable orbit at 300 km above Earth.



8-6

Solve ln (2x – 4)3 = 6.

ln (2x – 4)3 = 6

3 ln (2x – 4) = 6 Power Property

ln (2x – 4) = 2 Divide each side by 3.

2x – 4 = e2 Rewrite in exponential form.

x = Solve for x.e2 + 42

x 5.69 Use a calculator.

Check: ln (2 • 5.69 – 4)3 6 ln 401.95 6 5.996 6



8-6

Use natural logarithms to solve 4e3x + 1.2 = 14.

4e3x + 1.2 = 14

4e3x = 12.8 Subtract 1.2 from each side.

e3x = 3.2 Divide each side by 4.

ln e3x = ln 3.2 Take the natural logarithm of each side.

3x = ln 3.2 Simplify.

x = Solve for x.ln 3.23

x 0.388



8-6

An initial investment of $200 is now valued at $254.25. The

interest rate is 6%, compounded continuously. How long has the

money been invested?

A = Pert Continuously compounded interest formula.

254.25 = 200e0.06t Substitute 254.25 for A, 200 for P, and 0.06 for r.

1.27125 = e0.06t Divide each side by 200.

ln 1.27125 = ln e0.06t Take the natural logarithm of each side.

ln 1.27125 = 0.06t Simplify.

The money has been invested for 4 years.

= t Solve for t.ln 1.271250.06

4 t Use a calculator.



8-6



8-6


1. In 125

2. In 18

3. In 4

4. In 40,960

5. In

6. In 1

7. In

8. In

9. In

10. 20.92

1 81

m5

n3

xy

z4

3

a c

b2

23. 2.890

24. 1.151

25. 2.401

26. 5.493

27. 1.242

28. 23.752

29. 6 years

30. 78%

31. 1

32. 2

33. 10

34. 10

11. 24.13

12. 7.79 km/s; yes

13. 25 seconds

14. 134.476

15. 0.135

16. 1.078 1015

17. 1488.979

18. 5.482, –3.482

19. ±11.588

20. 110.196

21. ±2.241

22. ±0.908



8-6

35. 0

36.

37. 1

38. 83

39. 10.8

40. 301 days; 26 days

41. sometimes

42. never

43. always

44. about 5.8% per hour

45. 19.8 hours

46. about 40,000 bacteria

14

47. 3.6

48. 6.7

49. 9.4

50. 11.8

51. 13.9

52. 15.8

53. 17.5

54. 19.1

55. 542.31

56. 1

57. 0.0794

58. 81.286

59. 1.2639

60. no solution

61. 27,347.9

62. 78.342

63. No; using the Change of Base Formula would result in one of the log expressions being written as a quotient of logs, which couldn’t then be combined with the other expression to form a single logarithm.



8-6

64. a. y = 300e0.241t

b. 2002

c. 2006

d. t = , where y is thenumber of Internet users in million and t is time in years.

e. Substitute the number of users found in (b) and (c) into the equation in (d). Determine whether your answers in years are the same as t for each.

y

300ln ( ) 0.241

65. a. about 43 min

b. t ln

1.7, 6.0, 11.3, 18.1, 27.6, 43.1, 97.6

66. Check students’ work.

67. C

68. H

10.041

T 72 164



8-6

69. B

70. [4] a. 8%

b. A = Pert

= 500e0.08t

c. 1800 = 500e0.08t

3.6 = e0.08t

ln 3.6 = 0.08t

= t

16 tabout 16 years

[3] correct model computation error in (b) or (c)

[2] incorrect model, solved correctly

[1] correct model, but without work shown in (c)

ln 3.6 0.08

71. 4

72. 2.846

73. 0.272

74. 3333.3

75. 1.002

76. 9.0 10–5

77. y = ; yes

78. y = ;

yes

79. y = ± 5 – x ; no80. 4060 possible

combinations

x – 7 5

x – 10 2

3

1. Write 4 ln 6 – 2 ln 3 as a single natural logarithm.

2. Solve e3x = 15.

3. Simplify ln e7.

ln 144

about 0.903

7



8-6

2. 3.1.

ALGEBRA 2 CHAPTER 8ALGEBRA 2 CHAPTER 8

Exponential and Logarithmic FunctionsExponential and Logarithmic Functions

8-A

Page 472



8-A

4. 5. Answers may vary. Both answers should be of the form y = abx. For the growth model, b > 1. For the decay model, 0 < b < 1.

6. y = (3)x

7. y = 2(5)x

8. y = 4(4)x

9. 4.11 years

10. y = 3x translated up 2 units

11. y = 3( )x translated left 1 unit

13

12. y = 5x translated right 2 units and down 1 unit

13. y = 2x translated left 2 units and reflected over the x-axis

14. 3

15. 1

16. 2

17. 3

18. 0

19. 412

20.

21.

22.

23.

24. log2 2916

25. log

26. log7 a – log7 b

27. log 3 + 3 log x + 2 log y

28. 1

29. 13

30. –1

31. 3

a3

b2

32. Answers may vary.

Sample: Taking common

logarithms of both sides

gives 2x = . Taking

logarithms to the base 3 of

both sides gives

2x = log3 4, which by the

Change of Base Formula is

equivalent to 2x = .

33. –15.28

34. 3.89

35. 250

36. 0.01

log 4log 3

log 4log 3



8-A

37.

38.

39.

40. 1.359

41. 7.456

42. 17.058, –16.058

43. 4.780

44. 99 cm

log 16log 3

1log 2log 8log 7



8-A

Exploring Exponential Models

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