1. Exploring Exponential Models
ALGEBRA 2 LESSON 8-1
(For help, go to Lesson 1-2.)
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
for x = 0
6. 2x for x = –2 1 2 x
8-1

2. Exploring Exponential Models
ALGEBRA 2 LESSON 8-1
Solutions
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
for x = 0: = 1
6. 2x for x = –2: 2–2 = = = 1 2 1 2 x 1 22 1
2 • 2 1 4
8-1

3. Exploring Exponential Models
ALGEBRA 2 LESSON 8-1
Graph y = 3x.
x 3x y
–3 3–3 = .037
–2 3–2 = .1
–1 3–1 = .3
0 30 1
1 31 3
2 32 9 1 27 9 3
Step 1: Make a table of values.
Step 2: Graph the coordinates.
Connect the points with a
smooth curve.
8-1

4. Exploring Exponential Models
ALGEBRA 2 LESSON 8-1
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
= Substitute 0.7%, or for r.
= Simplify.
b. Suppose the rate of growth had continued to be 0.7%.
Write a function to model this population growth.
Relate: The population increases exponentially, so y = abx
Define: Let x = number of years after 1994.
Let y = the population (in millions).
8-1

5. Exploring Exponential Models
ALGEBRA 2 LESSON 8-1
(continued)
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.
8-1

6. Exploring Exponential Models
ALGEBRA 2 LESSON 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. 6 b
2 = b0 Substitute for x and y using (0, 2) and for a using . 6 b
2 = • 1 Any number to the zero power equals 1.
2 = Simplify.
b = 3 Solve for b. 6 b
8-1

7. Exploring Exponential Models
ALGEBRA 2 LESSON 8-1
(continued)
a = Use your equation for a. 6 b
a = Substitute 3 for b.
a = 2 Simplify.
y = 2 • 3x Substitute 2 for a and 3 for b in y = abx. 6 3
The exponential for a graph that includes (1, 6) and (0, 2) is y = 2 • 3x.
8-1

8. Exploring Exponential Models
ALGEBRA 2 LESSON 8-1
Without graphing, determine whether the function y = represents exponential growth or decay. 2 3 x
In y = , b = . 3 2 x
Since b < 1, the function represents exponential decay.
8-1

9. Exploring Exponential Models
ALGEBRA 2 LESSON 8-1
Graph y = 36(0.5)x. Identify the horizontal asymptote.
Step 1: Make a table of values.
x –3 –2 – y 1 2
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

10. Exploring Exponential Models
ALGEBRA 2 LESSON 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

11. Exploring Exponential Models
ALGEBRA 2 LESSON 8-1
(continued)
Write: y = ab x
    11,800 = a(0.8)0 Substitute using (0, 11,800).
     11,800 = a Solve for a.
y = 11,800(0.8)x Substitute a and b into y = abx.
y = 11,800(0.8)6   Evaluate for x = 6.
         Simplify.
The car’s depreciated value after 6 years will be about $3,090.
8-1

12. Exploring Exponential Models
ALGEBRA 2 LESSON 8-1 1. 2. 3. 4. 6. 7. 5. 8.
9. a
b. y = 6.08(1.0126)x, where x = 0 corresponds to 2000
pages 426–429 Exercises
8-1

13. Exploring Exponential Models
ALGEBRA 2 LESSON 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
18. exponential growth
19. exponential decay
20. exponential decay
21. exponential growth
22. exponential growth
23. exponential decay 1 3
24–31. y = 0 is the horizontal
asymptote.
24.
25.
26.
27.
28.
29. x
8-1

14. Exploring Exponential Models
ALGEBRA 2 LESSON 8-1
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 animals c.
about 47 years
47. 6
30.
31.
32. y = 100(0.5)x;
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 $
36. a. Tokyo: 26,444,000,
Mexico City:
18,850,649,
Bombay: 23,148,579,
São Paulo: 19,496,367
b. yes; Tokyo, Bombay, São Paulo, Mexico City
37. 63% increase
38. 30% increase
39. 35% decrease
40. 70% increase
% decrease
42. 75% decrease
43. a. 5.6%
b %
8-1

15. Exploring Exponential Models
ALGEBRA 2 LESSON 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
53. 2
54. y = 34(1.26)x, where x
represents 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

16. Exploring Exponential Models
ALGEBRA 2 LESSON 8-1
64. [4] y = abx
54 = ab2
a =
2 =  b
2 =
2b = 54
b = 27
b = 27
b = 9
y =  9x
[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.
67.
68. 6n n
69. 84r r 2
70.
71. Answers may vary.
Sample:
y = x3 – 5x2 + 4x
72. Answers may vary.
y = x3 – 7x – 6 54 b2 54 b2 1 2 54 b 3 2 3 3 2 3 2
2 x 3x 2 3 54 92 2 3 2 3
8-1

17. Exploring Exponential Models
ALGEBRA 2 LESSON 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 3 2 1
8-1

18. Exploring Exponential Models
ALGEBRA 2 LESSON 8-1
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 the
given points.
5. (0, 0.5), (1, 3) 6. (–1, 5), (0.5, 40) 1 4 x
y = 0
y = 0
growth
decay
y = 0.5(6)x
y = 20(4)x
8-1

19. Properties of Exponential Functions
ALGEBRA 2 LESSON 8-2
(For help, go to Lesson 2-6, 5-3, and 7-4.)
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 variables
are 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. 5 4 – x 1 7 6 –7
8-2

20. Properties of Exponential Functions
ALGEBRA 2 LESSON 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 5 4 – 1 x5 7 6
Solutions –7
8-2

21. Properties of Exponential Functions
ALGEBRA 2 LESSON 8-2
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 3 8 4 1 2 –
Step 1: Make a table of values
Step 2: Graph each function.
8-2

22. Properties of Exponential Functions
ALGEBRA 2 LESSON 8-2
Graph y = and y = – 2. 1 2 x 1 2
x–3
Step 1: Graph y = The horizontal asymptote is y = 0. 1 2 x
So shift the graph of the present function 3 units right and 2 units down.
Step 2: For y = – 2, h = 3 and k = –2. 1 2
x–3
The horizontal asymptote is y = –2.
8-2

23. Properties of Exponential Functions
ALGEBRA 2 LESSON 8-2
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. 1 2 1 6
Define: Let y = the amount of technetium-99m.
Let x = the number of hours elapsed.
Then x = the number of half-lives.
8-2

24. Properties of Exponential Functions
ALGEBRA 2 LESSON 8-2
(continued)
Write: y = 50
y = 50 Substitute 25 for x. 1 6
• 25 2 x
= Simplify.
2.784 1 2
4.16
After 25 hours, about mg of technetium-99m remains.
8-2

25. Properties of Exponential Functions
ALGEBRA 2 LESSON 8-2
Graph y = ex. Evaluate e3 to four decimal places.
Step 1: Graph y = ex.
Step 2: Find y when x = 3.
The value of e3 is about
8-2

26. Properties of Exponential Functions
ALGEBRA 2 LESSON 8-2
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, for r, and 3 for t.
= 100 • e Simplify.
= 100 • ( ) Evaluate e0.144.
Simplify.
You will have about $ in the account after three years.
8-2

27. Properties of Exponential Functions
ALGEBRA 2 LESSON 8-2
pages 434–436  Exercises
1–8. Asymptote is y = 0. 1. 2. 6. 7. 8. 3. 4. 5.
8-2

28. Properties of Exponential Functions
ALGEBRA 2 LESSON 8-2 9.
10.
11.
12.
13.
14.
15. y = ; 0.85 mg
16. y = ; 0.43 mg
17. y = ; 0.64 mg
21. 1
24. $
25. $448.30
26. $
27. 0 1 2
14.3
8.14
5730 x
8-2

29. Properties of Exponential Functions
ALGEBRA 2 LESSON 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. $
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 = ;
y =
33. y = –3x;
y = –3x–8 + 2 1 2
34. y = (2)x;
y = (2)x–6 – 7
35. y = – ;
y = – – 1
pascals yr
38. 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. 3 x
x + 4
x + 5
8-2

30. Properties of Exponential Functions
ALGEBRA 2 LESSON 8-2
39. a. GDP = 8.511(1.038)t where t = 0 corresponds to and is trillions.
b. It almost doubles. (about 195.7%)
c. In 9 years, the growth is about 40%.
40. a. $
b. $3.15 more
41. $399.97
42. exponential growth
43. exponential growth
44. exponential decay
45. exponential growth
46. exponential decay
47. exponential growth
48. a. y = 8001 – 3x, where y is the number of uninfected people and x represents days.
b 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

31. Properties of Exponential Functions
ALGEBRA 2 LESSON 8-2
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
58. y = –5(2)x
59. y = 8(0.5)x
60. y = 70(10)x
61. y = 10,000(0.8)x
63. –
64. 5( )
65. 2( ) –
70. x – 3, R –1 3 4
50. a m3
b. V = 2928 – 15(2x – 1)
c. ninth weekend
51. A
52. I
53. C
54. F
55. [2] A = Pert
8000 = Pe(0.06)(4)
= P
P = $
[1] answer only, without work shown
56. y = 3x
57. y = –2(4)x
8000
e(0.06)(4)
8-2

32. Properties of Exponential Functions
ALGEBRA 2 LESSON 8-2
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

33. Properties of Exponential Functions
ALGEBRA 2 LESSON 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.
about $2324 1 3
1.3956
8-2

34. Logarithmic Functions as Inverses
ALGEBRA 2 LESSON 8-3
(For help, go to Lessons 7-1 and 7-7.)
Solve each equation.
1. 8 = x3 2. x = = 3x = 43x
Graph each relation and its inverse on a coordinate plane.
5. y = 5x 6. y = 2x2 7. y = –x3 8. y = x 1 4 2
8-3

35. Logarithmic Functions as Inverses
ALGEBRA 2 LESSON 8-3
2. x = 2; since 16 = = 2, x = 16
6. y = 2x2
Inverse:
x = 2y2
y2 =
y2 = ±
8. y = x
x = y
y = 2x x 2
1. 8 = x3; since 23 = 8, x = 2
3. 27 = 3x; since 33 = 27, x = 3
5. y = 5x
x = 5y
y = x
7. y = –x3
x = –y3
y3 = –x
y = 1 5 3 –x 4 16
= 43x; since 6 = 3(2), x = 2
Solutions
8-3

36. Logarithmic Functions as Inverses
ALGEBRA 2 LESSON 8-3
Compare the amount of energy released in an earthquake that registers 6 on the Richter scale with one that registers 3.
Write a ratio.
E • 306
E • 303
= Simplify.
306
303
= 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.
8-3

37. Logarithmic Functions as Inverses
ALGEBRA 2 LESSON 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

38. Logarithmic Functions as Inverses
ALGEBRA 2 LESSON 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

39. Logarithmic Functions as Inverses
ALGEBRA 2 LESSON 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?
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
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

40. Logarithmic Functions as Inverses
ALGEBRA 2 LESSON 8-3
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

41. Logarithmic Functions as Inverses
ALGEBRA 2 LESSON 8-3
Graph y = log5 (x – 1) + 2.
Step 1: Make a table of values for the parent function.
Step 2: Graph the function by shifting the points from the table to the right 1 unit and up 2 units. 1
125 25 5 –3 –2 –1 x
y = log5 x
8-3

42. Logarithmic Functions as Inverses
ALGEBRA 2 LESSON 8-3
pages 441–444  Exercises
1. The earthquake in Missouri
released about 1.97 times
more energy.
2. The earthquake in Chile
released about 231 times
3. The earthquake in Missouri
released about 8,759,310
times more energy.
4. The earthquake in Missouri
released about 30 times
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 2 3 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
8-3

43. Logarithmic Functions as Inverses
ALGEBRA 2 LESSON 8-3
 10–6
 10–3
 10–8
 10–4
 10–7
 10–5
 10–4
 10–6
 10–4
35.
36.
37.
38.
39.
40.
; 0
42. –4.2147; –5
43. –1.0969; –2
; 2
45. –0.7782; –1
8-3

44. Logarithmic Functions as Inverses
ALGEBRA 2 LESSON 8-3
; 1
; 7
; 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
= 27
= 10–4
55. 16,807 = 75
56. 6 = 61
57. 1 = 40
= 3–2
= 2–1
= 101
= 213 9 2
8-3

45. Logarithmic Functions as Inverses
ALGEBRA 2 LESSON 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: to 10,242 years old, spear shaft: to 10,263 years old
b. 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
73.
74.
75. domain {x|x > 0}, range: all reals
76. domain {x|x > 0}, range: all reals
8-3

46. Logarithmic Functions as Inverses
ALGEBRA 2 LESSON 8-3
77. domain {x|x > 3}, range: all reals
78. domain {x|x > 0}, range: all reals
79. domain {x|x > 2}, range: all reals
80. domain {x|x > –1}, range: all reals
81. domain {x|x > 0}, range: all reals
82. domain {x|x > 0}, range: all reals
83. domain {x|x > t }, range: all reals
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

47. Logarithmic Functions as Inverses
ALGEBRA 2 LESSON 8-3 w3 3
t 2 5 z8 n
x p 1 2
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)
98.
y = 0
99.
y = 9
8-3

48. Logarithmic Functions as Inverses
ALGEBRA 2 LESSON 8-3
Evaluate each logarithm.
1. log log10(–100) 3. log5 1 25 5
undefined –2
Write each equation in exponential form.
4. log 0.01 = –2 5. log327 = 3
6. Graph y = log6(x – 3) + 1.
10–2 = 0.01
33 = 27
8-3

49. Properties of Logarithms
ALGEBRA 2 LESSON 8-4
(For help, go to Lessons 8-3 and 1-2.)
Simplify each expression.
1. log24 + log28 2. log39 – log log216 ÷ log264
Evaluate each expression for x = 3.
4. x3 – x 5. x5 • x x3 + x2 x6 x9
8-4

50. Properties of Logarithms
ALGEBRA 2 LESSON 8-4
1. log24 = x     log28 = y
2x = 4 2y = 8
x = 2 y = 3
log24 + log28 = = 5
2. log39 = x     log327 = y
3x = 9 3y = 27
log39 – log327 = 2 – 3 = –1
3. log216 = x     log264 = y
2x = 16 2y = 64
x = 4 y = 6
log216 ÷ log264 = 4 ÷ 6 = = 4 6 2 3
4. x3 – x for x = 3: 33 – 3 = 27 – 3 = 24
5. x5 • x2 for x = 3: 35 • 32 = 3(5+2) = 37 = 2187
for x = 3: = 3(6–9) = 3–3 = =
7. x3 + x2 for x = 3: = = 36 x6 x9 36 99 1 33 27
Solutions
8-4

51. Properties of Logarithms
ALGEBRA 2 LESSON 8-4
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 y x2 y
Quotient Property: logb = logb x2 – logb y x2 y
Power Property: logb x2 – logb y = 2 logb x – logb y
8-4

52. Properties of Logarithms
ALGEBRA 2 LESSON 8-4
Write each logarithmic expression as a single logarithm.
a. log4 64 – log4 16
log4 64 – log4 16 = log4 Quotient Property 64 16
= log4 4 or 1 Simplify.
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

53. Properties of Logarithms
ALGEBRA 2 LESSON 8-4
Expand each logarithm.
a. log7 t u
log = log7 t – log7 u Quotient Property t u
b. log(4p3)
log(4p3) = log 4 + log p3 Product Property
= log log p Power Property
8-4

54. Properties of Logarithms
ALGEBRA 2 LESSON 8-4
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 l1 l0 l2
8-4

55. Properties of Logarithms
ALGEBRA 2 LESSON 8-4
(continued)
L1 – L2 = 10 log l1 l0 l2
– 10 log
Find the decrease in loudness L1 – L2.
= 10 log l1 l0
0.40l1
– 10 log
Substitute l2 = 0.40l1.
– 10 log 0.40 •
Product Property
= 10 log l1 l0
– 10 ( log log )
= 10 log l1 l0
– 10 log 0.40 – 10 log
Distributive Property
= –10 log 0.40
Combine like terms.
4.0
Use a calculator.
The decrease in loudness would be about 4 decibels.
8-4

56. Properties of Logarithms
ALGEBRA 2 LESSON 8-4
pages 449–451  Exercises
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 2 3 m4 n 5 2k xy z
20. log log7 x + log7 y + log7 z
21. log log4 x
22. log log m – 2 log n
23. log5 r – log5 s
24. 2 log log3 x
25. log log (2x – 3)
26. 2 log a + 3 log b – 4 log c
log log x – log y
log log8 a
29. log s + log 7 – 2 log t 1
8-4

57. Properties of Logarithms
ALGEBRA 2 LESSON 8-4 t s
30. –logb x
31. 9 dB dB
33. –2
34. 1
35. 6
36. 2
37. 2
38. 1
39. 1
40. –2
41. 1 1 2
51. –1.398
53. –0.7782 dB
58. True; log2 4 = 2 and log2 8 = 3.
59. False; log3 3 = log3 3 , not log3 . 3
42. The coefficient is missing in log4 s;
log = log4 = (log4 t – log4 s)
= log4 t – log4 s.
43. Answers may vary. Sample:
log 150 = log 15 + log 10
49. –0.0969
50. –0.6021
8-4

58. Properties of Logarithms
ALGEBRA 2 LESSON 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. x y
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. dB 7 3 1 8
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

59. Properties of Logarithms
ALGEBRA 2 LESSON 8-4
81. log log log r – log s
logb x + logb y – logb z
log4 x + log4 y – log4 z – 4 log4 w
log (x2 – 4) – 2 log (x + 3)
log x + log 2 – log y 1 2 3 5 7 4 4 2x
2 y z3 27 2
mxn p
x2 y3 z5 3 1 y z
3 x5
73. log3
74. logx
75. log
76. log4
77. logb
78. log
79. 3 log log x – 3 log 5
80. 3 log m – 4 log n + 2 log p
86. log3 x + log3 y – 6 log3 z
log7 (r + 9) – 2 log7 s – log7 t
88. v = logb N bv = N MN = bu • bv = bu+v logb MN = u + v logb MN = logb M + logb N 1 2 3
8-4

60. Properties of Logarithms
ALGEBRA 2 LESSON 8-4
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)
u = logb M (given)
2. bu = M (Rewrite in exponential form.)
3. v = logb N (given)
4. bv = N (Rewrite in exponential form.)
= = 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) M N bu bv
91. B
92. G
93. D
94. [2] By the Quotient Property, log = log – = –
[1] correct answer, without work shown 1 2 10 20
8-4

61. Properties of Logarithms
ALGEBRA 2 LESSON 8-4
101.
102. 2
103. –
104. i, –4i
105. –2i, –4 – i
106. – 2,– i – 1
107.
108. 2i + 3, –i 64 7
3, 5
7, 11
95. [4] 1) log = log 24 – log 2; Quotient Property
2) log 2 • 6 = log log 6; Product Property
3) log =
log 144; Power Property
4) log 3 • 22 = log log 2; Product and Power Properties
[3] log 12 written three ways 24 2 1
[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 1 4 2 3
125
8-4

62. Properties of Logarithms
ALGEBRA 2 LESSON 8-4
Write each expression as a single logarithm. State the property you used.
1. log 12 – log log115 + log117
Expand each logarithm.
3. logc 4. log3x4
Use the properties of logarithms to evaluate each expression.
5. log log logyy
log 4; Quotient Property
log11(53 • 7); Power Property and Product Property a b
4 log3x
logca – logcb 1 2 –1 1 2
8-4

63. Exponential and Logarithmic Equations
ALGEBRA 2 LESSON 8-5
(For help, go to Lessons 8-3 and 7-4.)
Evaluate each logarithm.
1. log981 • log93 2. log10 • log39
3. log216 ÷ log28 4. Simplify 2 3 –
8-5

64. Exponential and Logarithmic Equations
ALGEBRA 2 LESSON 8-5
= = = = = 1 3
(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 4
2. log10 = x     log39 = y
10x = 10 3y = 9
x = 1 y = 2
log10 • log39 = 1 • 2 = 2
125 52 25
( )
Solutions –
8-5

65. Exponential and Logarithmic Equations
ALGEBRA 2 LESSON 8-5
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 16
2 log 5
Use a calculator.
Check: 52x
52(0.8614) 16
8-5

66. Exponential and Logarithmic Equations
ALGEBRA 2 LESSON 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 12
log 6
1.387 Use a calculator.
1.0792
0.7782
log6 12 = log3 x Write an equation.
log3x Substitute log6 12 =
1.387 Use the Change of Base Formula.
log x
log 3
8-5

67. Exponential and Logarithmic Equations
ALGEBRA 2 LESSON 8-5
(continued)
1.387 • log 3 log x Multiply each side by log 3.
1.387 • log x Use a calculator.
log x Simplify.
x Write in exponential form.
Use a calculator.
The expression log6 12 is approximately equal to , or log
8-5

68. Exponential and Logarithmic Equations
ALGEBRA 2 LESSON 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 Use a calculator to solve for x.
8-5

69. Exponential and Logarithmic Equations
ALGEBRA 2 LESSON 8-5
Solve 43x = 1100 by graphing.
Graph the equations y = 43x and y = 1100.
Find the point of intersection.
The solution is x
8-5

70. Exponential and Logarithmic Equations
ALGEBRA 2 LESSON 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 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
Pop. P(t)
Step 1: Enter the data into your calculator.
Step 2: Use the Exp Reg feature to find the exponential function that fits the data.
8-5

71. Exponential and Logarithmic Equations
ALGEBRA 2 LESSON 8-5
(continued)
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.
8-5

72. Exponential and Logarithmic Equations
ALGEBRA 2 LESSON 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,
log (2 • 5001 – 2) 4
Check: log (2x – 2)
8-5

73. Exponential and Logarithmic Equations
ALGEBRA 2 LESSON 8-5
Solve 3 log x – log 2 = 5.
3 log x – log 2 = 5 x3 2
Log ( ) = 5 Write as a single logarithm. x3 2
= 105 Write in exponential form.
x3 = 2(100,000) Multiply each side by 2.
x = , or about 3
The solution is , or about 3
8-5

74. Exponential and Logarithmic Equations
ALGEBRA 2 LESSON 8-5
pages 456–460 Exercises
5. 3
; log8 729
; log
; log
; log
; log
; log8 343
; log
; log
8-5

75. Exponential and Logarithmic Equations
ALGEBRA 2 LESSON 8-5
31. about 7.1 years
32. about 2018
, or
35. 33
36. 10,000
, or
38. 12
, or
– 1, or
41. 2
42. 3  108
, , or ,606.8
44. 5
45.
47. 7 1 60 1 4 10
8-5

76. Exponential and Logarithmic Equations
ALGEBRA 2 LESSON 8-5
59. 3
60. –
61. a. Florida growth factor = , y = 15,982,378 • (1.0213)x; New York growth factor = , y = 18,976,457 • (1.0054)x
b. 2011
48. a b
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.
= 8(1.2)x, 5 years
51. 2 = 10( ) , 2.7 min
= 125(0.88)x,
4 years
53. –1
54. 3
55.
56. 3
57.
58. –2 1 2 x
1.17 3
8-5

77. Exponential and Logarithmic Equations
ALGEBRA 2 LESSON 8-5
65. Answers may vary. Sample: log x = = x, x
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) ; using this model, there will be 10,000 manatees in about 2015.
62. a. Texas growth factor = , y = 20,851,820 (1.0208)x; California growth factor = , 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.
– 1
log102
log101 = /
8-5

78. Exponential and Logarithmic Equations
ALGEBRA 2 LESSON 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 %
78. a. 10–3 W/m2, 106 W/m2
b. 109 times more intense
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 b
log a
68.
69.
70.
71.
72.
73.
74.
75.
log 2
log 7
log 8
log 3
log 140
log 5
log 3.3
log 9
log 3x
log 4
log (1 – x)
log 6
log x
log (x + 1)
8-5

79. Exponential and Logarithmic Equations
ALGEBRA 2 LESSON 8-5
84. 10
91. 1
97. a. bassoon, guitar, harp, violin, viola, cello
b. bassoon, guitar, harp, cello, bass
c. harp, violin
d. harp, violin
,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 x
(0.5)5430
8-5

80. Exponential and Logarithmic Equations
ALGEBRA 2 LESSON 8-5
113. log log x – 2 log y
114. log3 x – log3 y
log log2 x
116. log log3 (2x – 3)
117. log 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
,031 m above sea level
105. a. 91 hours or 4 days
b mg or mg
c. Estimate in hours is more
accurate; the days have a
larger rounding error. 1 2
8-5

81. Exponential and Logarithmic Equations
ALGEBRA 2 LESSON 8-5
123. ±2, ±2i
124. ± 2, ± 3
125. –2, –1, 3
126. 2(x) + 2(x + 12) = 128; 26 ft
= 133; 119
679 + x 6
8-5

82. Exponential and Logarithmic Equations
ALGEBRA 2 LESSON 8-5
Use mental math to solve each equation.
1. 2x = 2. log42 = x x = 1
4. Solve 52x = 125. 1 8 1 2 –3 3 2
8-5

83. Natural Logarithms
ALGEBRA 2 LESSON 8-6
(For help, go to Lessons 8-2 and 8-5.)
Use your calculator to evaluate each expression to the nearest thousandth.
1. e5 2. 2e3 3. e– e 1 e
Solve.
6. log3x = 4 7. log164 = x 8. log16x = 4
8-6

84. Natural Logarithms 7. log164 = x 16x = 4 x =
ALGEBRA 2 LESSON 8-6
7. log164 = x
16x = 4
x = 1 2
1. e e
3. e–
5. 4.2e e
6. log3x = 4
34 = x
81 = x
8. log16x = 4
164 = x
65,536 = x
Solutions
8-6

85. Write 2 ln 12 – ln 9 as a single natural logarithm.
Natural Logarithms
ALGEBRA 2 LESSON 8-6
Write 2 ln 12 – ln 9 as a single natural logarithm.
2 ln 12 – ln 9 = ln 122 – ln 9 Power Property
= ln Quotient Property
122 9
= ln 16 Simplify.
8-6

86. v = –0.0098t + c ln R Use the formula.
Natural Logarithms
ALGEBRA 2 LESSON 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) ln 22 Substitute.
– (3.091) Use a calculator.
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

87. 3 ln (2x – 4) = 6 Power Property
Natural Logarithms
ALGEBRA 2 LESSON 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 + 4 2
x Use a calculator.
Check: ln (2 • 5.69 – 4) ln
8-6

88. Use natural logarithms to solve 4e3x + 1.2 = 14.
ALGEBRA 2 LESSON 8-6
Use natural logarithms to solve 4e3x = 14.
4e3x = 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 3.2 3 x
8-6

89. A = Pert Continuously compounded interest formula.
Natural Logarithms
ALGEBRA 2 LESSON 8-6
An initial investment of $200 is now valued at $ The interest rate is 6%, compounded continuously. How long has the money been invested?
A = Pert Continuously compounded interest formula.
= 200e0.06t Substitute for A, 200 for P, and 0.06 for r.
= e0.06t Divide each side by 200.
ln = ln e0.06t Take the natural logarithm of each side.
ln = 0.06t Simplify.
= t Solve for t. ln
0.06
4 t Use a calculator.
The money has been invested for 4 years.
8-6

90. Natural Logarithms pages 464–467 Exercises 1. In 125 2. In 18 3. In 4
ALGEBRA 2 LESSON 8-6
pages 464–467 Exercises
1. In 125
2. In 18
3. In 4
4. In 40,960
5. In
6. In 1
7. In
8. In
9. In 1 81 m5 n3 xy z4 3
a c b2
29. 6 years
30. 78%
31. 1
32. 2
33. 10
34. 10
km/s; yes
seconds
 1015
, –3.482
19. ±11.588
21. ±2.241
22. ±0.908
8-6

91. 63. No; using the Change of Base Formula would result in one of
Natural Logarithms
ALGEBRA 2 LESSON 8-6
35. 0
36.
37. 1
38. 83
days; 26 days
41. sometimes
42. never
43. always
44. about 5.8% per hour
hours
46. about 40,000 bacteria 1 4
56. 1
60. no solution
61. 27,347.9
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

92. ( ) Natural Logarithms 64. a. y = 300e0.241t 65. a. about 43 min
ALGEBRA 2 LESSON 8-6
64. a. y = 300e0.241t
b. 2002
c. 2006
d. t = , where y is the
number 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
300 ln
( )
0.241
65. a. about 43 min
b. t  ln
1.7, 6.0, 11.3, 18.1, , 43.1, 97.6
66. Check students’ work.
67. C
68. H 1
0.041
T  72
164
8-6

93. [3] correct model computation error in (b) or (c)
Natural Logarithms
ALGEBRA 2 LESSON 8-6
69. B
70. [4] a. 8%
b. A = Pert
= 500e0.08t
c = 500e0.08t
3.6 = e0.08t
ln 3.6 = 0.08t
= t t
about 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
 10–5
77. y = ; yes
78. y = ; yes
79. y = ± – x ; no
possible combinations
x – 7 5
x – 10 2 3
8-6

94. 1. Write 4 ln 6 – 2 ln 3 as a single natural logarithm.
Natural Logarithms
ALGEBRA 2 LESSON 8-6
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

95. Exponential and Logarithmic Functions
ALGEBRA 2 CHAPTER 8
Page 472 1. 2. 3.
8-A

96. Exponential and Logarithmic Functions
ALGEBRA 2 CHAPTER 8 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
years
10. y = 3x translated up 2 units
11. y = 3( )x translated left 1 unit 1 3
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. 4 2
8-A

97. Exponential and Logarithmic Functions
ALGEBRA 2 CHAPTER 8
20.
21.
22.
23.
24. log2 2916
25. log
26. log7 a – log7 b
27. log log x log y
28. 1
29. 13
30. –1
31. 3
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 a3 b2
log 4
log 3
log 4
log 3
8-A

98. Exponential and Logarithmic Functions
ALGEBRA 2 CHAPTER 8
37.
38.
39.
, –16.058 cm
log 16
log 3 1
log 2
log 8
log 7
8-A