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
ALGEBRA 2 LESSON 8-1ALGEBRA 2 LESSON 8-1
Exploring Exponential ModelsExploring Exponential Models
8-1
x 12
0
Graph y = 3x.
Step 2: Graph the coordinates. Connect the points with a smooth curve.
ALGEBRA 2 LESSON 8-1ALGEBRA 2 LESSON 8-1
Exploring Exponential ModelsExploring Exponential Models
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.
ALGEBRA 2 LESSON 8-1ALGEBRA 2 LESSON 8-1
Exploring Exponential ModelsExploring Exponential Models
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.
ALGEBRA 2 LESSON 8-1ALGEBRA 2 LESSON 8-1
Exploring Exponential ModelsExploring Exponential Models
(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
ALGEBRA 2 LESSON 8-1ALGEBRA 2 LESSON 8-1
Exploring Exponential ModelsExploring Exponential Models
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
ALGEBRA 2 LESSON 8-1ALGEBRA 2 LESSON 8-1
Exploring Exponential ModelsExploring Exponential Models
(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.
ALGEBRA 2 LESSON 8-1ALGEBRA 2 LESSON 8-1
Exploring Exponential ModelsExploring Exponential Models
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.
ALGEBRA 2 LESSON 8-1ALGEBRA 2 LESSON 8-1
Exploring Exponential ModelsExploring Exponential Models
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.
ALGEBRA 2 LESSON 8-1ALGEBRA 2 LESSON 8-1
Exploring Exponential ModelsExploring Exponential Models
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.
ALGEBRA 2 LESSON 8-1ALGEBRA 2 LESSON 8-1
Exploring Exponential ModelsExploring Exponential Models
(continued)
8-1
ALGEBRA 2 LESSON 8-1ALGEBRA 2 LESSON 8-1
Exploring Exponential ModelsExploring Exponential Models
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
ALGEBRA 2 LESSON 8-1ALGEBRA 2 LESSON 8-1
Exploring Exponential ModelsExploring Exponential Models
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
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%
ALGEBRA 2 LESSON 8-1ALGEBRA 2 LESSON 8-1
Exploring Exponential ModelsExploring Exponential Models
8-1
ALGEBRA 2 LESSON 8-1ALGEBRA 2 LESSON 8-1
Exploring Exponential ModelsExploring Exponential Models
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.
ALGEBRA 2 LESSON 8-1ALGEBRA 2 LESSON 8-1
Exploring Exponential ModelsExploring Exponential Models
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
ALGEBRA 2 LESSON 8-1ALGEBRA 2 LESSON 8-1
Exploring Exponential ModelsExploring Exponential Models
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
ALGEBRA 2 LESSON 8-1ALGEBRA 2 LESSON 8-1
Exploring Exponential ModelsExploring Exponential Models
8-1
x
(For help, go to Lesson 2-6, 5-3, and 7-4.)
ALGEBRA 2 LESSON 8-2ALGEBRA 2 LESSON 8-2
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
ALGEBRA 2 LESSON 8-2ALGEBRA 2 LESSON 8-2
Properties of Exponential FunctionsProperties of Exponential Functions
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.
ALGEBRA 2 LESSON 8-2ALGEBRA 2 LESSON 8-2
Properties of Exponential FunctionsProperties of Exponential Functions
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.
ALGEBRA 2 LESSON 8-2ALGEBRA 2 LESSON 8-2
Properties of Exponential FunctionsProperties of Exponential Functions
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
ALGEBRA 2 LESSON 8-2ALGEBRA 2 LESSON 8-2
Properties of Exponential FunctionsProperties of Exponential Functions
8-2
After 25 hours, about 2.784 mg of technetium-99m remains.
ALGEBRA 2 LESSON 8-2ALGEBRA 2 LESSON 8-2
Properties of Exponential FunctionsProperties of Exponential Functions
(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.
ALGEBRA 2 LESSON 8-2ALGEBRA 2 LESSON 8-2
Properties of Exponential FunctionsProperties of Exponential Functions
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.
ALGEBRA 2 LESSON 8-2ALGEBRA 2 LESSON 8-2
Properties of Exponential FunctionsProperties of Exponential Functions
8-2
pages 434–436 Exercises
1–8. Asymptote is y = 0.
1.
2.
3.
4.
5.
6.
7.
8.
ALGEBRA 2 LESSON 8-2ALGEBRA 2 LESSON 8-2
Properties of Exponential FunctionsProperties of Exponential Functions
8-2
ALGEBRA 2 LESSON 8-2ALGEBRA 2 LESSON 8-2
Properties of Exponential FunctionsProperties of Exponential Functions
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
ALGEBRA 2 LESSON 8-2ALGEBRA 2 LESSON 8-2
Properties of Exponential FunctionsProperties of Exponential Functions
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
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. 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.
ALGEBRA 2 LESSON 8-2ALGEBRA 2 LESSON 8-2
Properties of Exponential FunctionsProperties of Exponential Functions
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
ALGEBRA 2 LESSON 8-2ALGEBRA 2 LESSON 8-2
Properties of Exponential FunctionsProperties of Exponential Functions
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
ALGEBRA 2 LESSON 8-2ALGEBRA 2 LESSON 8-2
Properties of Exponential FunctionsProperties of Exponential Functions
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
ALGEBRA 2 LESSON 8-2ALGEBRA 2 LESSON 8-2
Properties of Exponential FunctionsProperties of Exponential Functions
8-2
(For help, go to Lessons 7-1 and 7-7.)
ALGEBRA 2 LESSON 8-3ALGEBRA 2 LESSON 8-3
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
ALGEBRA 2 LESSON 8-3ALGEBRA 2 LESSON 8-3
Logarithmic Functions as InversesLogarithmic Functions as Inverses
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
ALGEBRA 2 LESSON 8-3ALGEBRA 2 LESSON 8-3
Logarithmic Functions as InversesLogarithmic Functions as Inverses
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.
ALGEBRA 2 LESSON 8-3ALGEBRA 2 LESSON 8-3
Logarithmic Functions as InversesLogarithmic Functions as Inverses
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.
ALGEBRA 2 LESSON 8-3ALGEBRA 2 LESSON 8-3
Logarithmic Functions as InversesLogarithmic Functions as Inverses
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.
ALGEBRA 2 LESSON 8-3ALGEBRA 2 LESSON 8-3
Logarithmic Functions as InversesLogarithmic Functions as Inverses
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.
ALGEBRA 2 LESSON 8-3ALGEBRA 2 LESSON 8-3
Logarithmic Functions as InversesLogarithmic Functions as Inverses
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
ALGEBRA 2 LESSON 8-3ALGEBRA 2 LESSON 8-3
Logarithmic Functions as InversesLogarithmic Functions as Inverses
8-3
ALGEBRA 2 LESSON 8-3ALGEBRA 2 LESSON 8-3
Logarithmic Functions as InversesLogarithmic Functions as Inverses
8-3
pages 441–444 Exercises
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
ALGEBRA 2 LESSON 8-3ALGEBRA 2 LESSON 8-3
Logarithmic Functions as InversesLogarithmic Functions as Inverses
8-3
ALGEBRA 2 LESSON 8-3ALGEBRA 2 LESSON 8-3
Logarithmic Functions as InversesLogarithmic Functions as Inverses
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
ALGEBRA 2 LESSON 8-3ALGEBRA 2 LESSON 8-3
Logarithmic Functions as InversesLogarithmic Functions as Inverses
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
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
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
ALGEBRA 2 LESSON 8-3ALGEBRA 2 LESSON 8-3
Logarithmic Functions as InversesLogarithmic Functions as Inverses
8-3
98.
y = 0
99.
y = 9
ALGEBRA 2 LESSON 8-3ALGEBRA 2 LESSON 8-3
Logarithmic Functions as InversesLogarithmic Functions as Inverses
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
ALGEBRA 2 LESSON 8-3ALGEBRA 2 LESSON 8-3
Logarithmic Functions as InversesLogarithmic Functions as Inverses
8-3
1 25
(For help, go to Lessons 8-3 and 1-2.)
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
ALGEBRA 2 LESSON 8-4ALGEBRA 2 LESSON 8-4
Properties of LogarithmsProperties of Logarithms
8-4
ALGEBRA 2 LESSON 8-4ALGEBRA 2 LESSON 8-4
Properties of LogarithmsProperties of Logarithms
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
ALGEBRA 2 LESSON 8-4ALGEBRA 2 LESSON 8-4
Properties of LogarithmsProperties of Logarithms
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).
ALGEBRA 2 LESSON 8-4ALGEBRA 2 LESSON 8-4
Properties of LogarithmsProperties of Logarithms
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
ALGEBRA 2 LESSON 8-4ALGEBRA 2 LESSON 8-4
Properties of LogarithmsProperties of Logarithms
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
ALGEBRA 2 LESSON 8-4ALGEBRA 2 LESSON 8-4
Properties of LogarithmsProperties of Logarithms
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.
ALGEBRA 2 LESSON 8-4ALGEBRA 2 LESSON 8-4
Properties of LogarithmsProperties of Logarithms
(continued)
8-4
ALGEBRA 2 LESSON 8-4ALGEBRA 2 LESSON 8-4
Properties of LogarithmsProperties of Logarithms
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
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
ALGEBRA 2 LESSON 8-4ALGEBRA 2 LESSON 8-4
Properties of LogarithmsProperties of Logarithms
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
ALGEBRA 2 LESSON 8-4ALGEBRA 2 LESSON 8-4
Properties of LogarithmsProperties of Logarithms
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.
=/
ALGEBRA 2 LESSON 8-4ALGEBRA 2 LESSON 8-4
Properties of LogarithmsProperties of Logarithms
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
ALGEBRA 2 LESSON 8-4ALGEBRA 2 LESSON 8-4
Properties of LogarithmsProperties of Logarithms
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
ALGEBRA 2 LESSON 8-4ALGEBRA 2 LESSON 8-4
Properties of LogarithmsProperties of Logarithms
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
ALGEBRA 2 LESSON 8-4ALGEBRA 2 LESSON 8-4
Properties of LogarithmsProperties of Logarithms
8-4
12
ab
Evaluate each logarithm.
1. log981 • log93 2. log10 • log39
3. log216 ÷ log28 4. Simplify 125 .
ALGEBRA 2 LESSON 8-5ALGEBRA 2 LESSON 8-5
(For help, go to Lessons 8-3 and 7-4.)
Exponential and Logarithmic EquationsExponential and Logarithmic Equations
8-5
2
3–
ALGEBRA 2 LESSON 8-5ALGEBRA 2 LESSON 8-5
Exponential and Logarithmic EquationsExponential and Logarithmic Equations
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
0.8614 Use a calculator.
Check: 52x 16
52(0.8614) 16
ALGEBRA 2 LESSON 8-5ALGEBRA 2 LESSON 8-5
Exponential and Logarithmic EquationsExponential and Logarithmic Equations
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
ALGEBRA 2 LESSON 8-5ALGEBRA 2 LESSON 8-5
Exponential and Logarithmic EquationsExponential and Logarithmic Equations
8-5
4.589 Use a calculator.
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.
ALGEBRA 2 LESSON 8-5ALGEBRA 2 LESSON 8-5
Exponential and Logarithmic EquationsExponential and Logarithmic Equations
(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.
ALGEBRA 2 LESSON 8-5ALGEBRA 2 LESSON 8-5
Exponential and Logarithmic EquationsExponential and Logarithmic Equations
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.
ALGEBRA 2 LESSON 8-5ALGEBRA 2 LESSON 8-5
Exponential and Logarithmic EquationsExponential and Logarithmic Equations
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.
ALGEBRA 2 LESSON 8-5ALGEBRA 2 LESSON 8-5
Exponential and Logarithmic EquationsExponential and Logarithmic Equations
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.
ALGEBRA 2 LESSON 8-5ALGEBRA 2 LESSON 8-5
Exponential and Logarithmic EquationsExponential and Logarithmic Equations
(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
ALGEBRA 2 LESSON 8-5ALGEBRA 2 LESSON 8-5
Exponential and Logarithmic EquationsExponential and Logarithmic Equations
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
ALGEBRA 2 LESSON 8-5ALGEBRA 2 LESSON 8-5
Exponential and Logarithmic EquationsExponential and Logarithmic Equations
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
pages 456–460 Exercises
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
ALGEBRA 2 LESSON 8-5ALGEBRA 2 LESSON 8-5
Exponential and Logarithmic EquationsExponential and Logarithmic Equations
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
ALGEBRA 2 LESSON 8-5ALGEBRA 2 LESSON 8-5
Exponential and Logarithmic EquationsExponential and Logarithmic Equations
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
ALGEBRA 2 LESSON 8-5ALGEBRA 2 LESSON 8-5
Exponential and Logarithmic EquationsExponential and Logarithmic Equations
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
ALGEBRA 2 LESSON 8-5ALGEBRA 2 LESSON 8-5
Exponential and Logarithmic EquationsExponential and Logarithmic Equations
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 =/
ALGEBRA 2 LESSON 8-5ALGEBRA 2 LESSON 8-5
Exponential and Logarithmic EquationsExponential and Logarithmic Equations
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
ALGEBRA 2 LESSON 8-5ALGEBRA 2 LESSON 8-5
Exponential and Logarithmic EquationsExponential and Logarithmic Equations
8-5
ALGEBRA 2 LESSON 8-5ALGEBRA 2 LESSON 8-5
Exponential and Logarithmic EquationsExponential and Logarithmic Equations
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
ALGEBRA 2 LESSON 8-5ALGEBRA 2 LESSON 8-5
Exponential and Logarithmic EquationsExponential and Logarithmic Equations
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
ALGEBRA 2 LESSON 8-5ALGEBRA 2 LESSON 8-5
Exponential and Logarithmic EquationsExponential and Logarithmic Equations
8-5
18
(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–2 4. 5. 4.2e
Solve.
6. log3x = 4 7. log164 = x 8. log16x = 4
ALGEBRA 2 LESSON 8-6ALGEBRA 2 LESSON 8-6
Natural LogarithmsNatural Logarithms
8-6
1e
ALGEBRA 2 LESSON 8-6ALGEBRA 2 LESSON 8-6
Natural LogarithmsNatural Logarithms
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.
ALGEBRA 2 LESSON 8-6ALGEBRA 2 LESSON 8-6
Natural LogarithmsNatural Logarithms
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.
ALGEBRA 2 LESSON 8-6ALGEBRA 2 LESSON 8-6
Natural LogarithmsNatural Logarithms
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
ALGEBRA 2 LESSON 8-6ALGEBRA 2 LESSON 8-6
Natural LogarithmsNatural Logarithms
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
ALGEBRA 2 LESSON 8-6ALGEBRA 2 LESSON 8-6
Natural LogarithmsNatural Logarithms
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.
ALGEBRA 2 LESSON 8-6ALGEBRA 2 LESSON 8-6
Natural LogarithmsNatural Logarithms
8-6
ALGEBRA 2 LESSON 8-6ALGEBRA 2 LESSON 8-6
Natural LogarithmsNatural Logarithms
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
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
ALGEBRA 2 LESSON 8-6ALGEBRA 2 LESSON 8-6
Natural LogarithmsNatural Logarithms
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.
ALGEBRA 2 LESSON 8-6ALGEBRA 2 LESSON 8-6
Natural LogarithmsNatural Logarithms
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
ALGEBRA 2 LESSON 8-6ALGEBRA 2 LESSON 8-6
Natural LogarithmsNatural Logarithms
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
ALGEBRA 2 LESSON 8-6ALGEBRA 2 LESSON 8-6
Natural LogarithmsNatural Logarithms
8-6
2. 3.1.
ALGEBRA 2 CHAPTER 8ALGEBRA 2 CHAPTER 8
Exponential and Logarithmic FunctionsExponential and Logarithmic Functions
8-A
Page 472
ALGEBRA 2 CHAPTER 8ALGEBRA 2 CHAPTER 8
Exponential and Logarithmic FunctionsExponential and Logarithmic Functions
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
ALGEBRA 2 CHAPTER 8ALGEBRA 2 CHAPTER 8
Exponential and Logarithmic FunctionsExponential and Logarithmic Functions
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
ALGEBRA 2 CHAPTER 8ALGEBRA 2 CHAPTER 8
Exponential and Logarithmic FunctionsExponential and Logarithmic Functions
8-A
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