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Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
(For help, go to Lessons 1-2 and 1-6.)
Simplify each expression.
1
1. 23
2. 42
3. 42  22
4. (–3)3
5. –33
6. 62  12
Evaluate each expression for a = 2, b = –1, c = 0.5.
7.
a
2a
8. bc
c
9.
8-1
ab
bc
Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
Solutions
1. 23 = 2 • 2 • 2 = 8
1 = 1 = 1
42
4 • 4 16
2
3. 42  22 = 42 = 4 • 4 = 16 = 4
2•2 4
2
2.
4. (–3)3 = (–3)(–3)(–3) = 9(–3) = –27
5. –33 = –(3 • 3 • 3) = –(9 • 3) = –27
6. 62  12 = 36  12 = 3
a for a = 2: 2 = 1
2•2 2
2a
8. bc for b = –1, c = 0.5: –1 • 0.5 = –1
c
0.5
9. ab for a = 2, b = –1, c = 0.5: 2 • (–1) = 2 = 4
bc
(–1) • 0.5 0.5
7.
8-1
Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
Simplify.
a.
1
3–2 = 32
1
= 9
b. (–22.4)0 = 1
Use the definition of negative
exponent.
Simplify.
Use the definition of zero as an
exponent.
8-1
Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
Simplify
a. 3ab –2 = 3a
1
b2
b.
Rewrite using a division symbol.
Use the definition of negative
exponent.
=
3a
b2
1
–3
=
1

x
–3
x
1
=1 3
x
Use the definition of negative
exponent.
Simplify.
= 1 • x3
1
Multiply by the reciprocal of x3 ,
which is x 3.
= x3
Identity Property of Multiplication
8-1
Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
Evaluate 4x 2y –3 for x = 3 and y = –2.
Method 1: Write with positive exponents first.
4x 2
4x 2y –3 = y 3
Use the definition of negative exponent.
4(3)2
=
(–2)3
36
Substitute 3 for x and –2 for y.
1
= –8 = –4 2
Simplify.
8-1
Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
(continued)
Method 2: Substitute first.
4x 2y –3 = 4(3)2(–2)–3
4(3)2
=
(–2)3
36
Substitute 3 for x and –2 for y.
Use the definition of negative exponent.
1
= –8 = –4 2
Simplify.
8-1
Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
In the lab, the population of a certain bacteria doubles every
month. The expression 3000 • 2m models a population of 3000
bacteria after m months of growth. Evaluate the expression for m = 0
and m = –2. Describe what the value of the expression represents in
each situation.
a. Evaluate the expression for m = 0.
3000 • 2m = 3000 • 20
= 3000 • 1
Substitute 0 for m.
Simplify.
= 3000
When m = 0, the value of the expression is 3000. This represents the
initial population of the bacteria. This makes sense because when m = 0,
no time has passed.
8-1
Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
(continued)
b. Evaluate the expression for m = –2.
3000 • 2m = 3000 • 2–2
Substitute –2 for m.
= 3000 • 1
4
= 750
Simplify.
When m = –2, the value of the expression is 750. This represents the 750
bacteria in the population 2 months before the present population of 3000
bacteria.
8-1
Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
pages 397–399 Exercises
1. –1
2. 1
16
3. 1
25
4. – 1
25
5. 1
16
6. – 1
81
7. 1
64
8. – 1
12
11. – 1
64
12. – 1
64
13. –2
14. 3; 4
15. 0; –3
16. –5
17. 3a
18. 54
x
19. x 7
9. 1
20. c
10. 1
21.
78
1
25p
22. 14
a
3
x2y
24. 7a
3b 2 w
25. 51 7
x y
7
26. y 5
x
23.
33. 1
25
34. 1
9
35. – 1
9
36. 1
37. 3
27. 4c3
38.
7st 3
28.
5
39.
29.
40.
6
ac 3
30. x 2
8z 7
31. y 7
t 11
32. 14
m 2t 5
8-1
41.
42.
25
1
100
25
81
81
25
– 25
27
1
25
43. –27
Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
44. – 27
54. 10–4
63. 45
45. a. $20.48; $.32
b. No; the value of the
allowance rapidly
becomes very great.
55. 10–5
64. 6
400
56. 0.001
57. 0.000001
65. 40
58. 0.7
66. 1
59. 0.03
67. – 1
47. pos.
60. 0.0005
68. 16
48. pos.
61. a. 5–2, 5–1, 50, 51, 52
b. 54
n
c. a
46. neg.
49. neg.
50. neg.
51. 10–1
52. 10–2
53. 10–3
4
1
62. In –30, 3 is raised to the
zero power, and then the
opposite is determined. In
(–3)0, the number –3 is
raised to the zero power.
8-1
243
69. 2
9
70. 1
8
71. 1
16
72. –1
Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
73.
74. a.
b.
79. a. 1 correct, 0.4096; 2 correct, 0.1536;
3 correct, 0.0256; 4 correct, 0.0016
b. 0 or 1
1
They are reciprocals for a =/ 0;
1 = a –n and 1 = 1 = an.
an
a –n 1
an
75. A, B, D
76. Check students’ work.
77. No;
•
= 9 • = 9.
The product of reciprocals
should be 1.
3x–2
3x2
x0
78. The student multiplied b by zero
instead of raising b to the zero
power, which would equal 1.
80. about 4 students; about 16 students;
about 29 students
81. 8 – 48m2
82. 21
83. 141
m2
84. 2.9375
4
nr 7y 2
86. –7 1
4
85.
87. 1 and –1
8-1
Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
88. 2
3
89. 1
9
90. 1
64
95.
c. Answers may vary slightly.
Sample: y = 53x – 4328
d. Answers may vary slightly.
Sample: $1,237,000,000
96.
98. y = –x + 4
91. 1
99. y = 5x – 2
92. 0.26
100. y = 2 x – 3
93. 258.4
94.
5
101. y = – 3 x – 17
11
97. a-b.
102. y = 5 x + 1
9
3
103. y = 1.25x – 3.79
8-1
Zero and Negative Exponents
ALGEBRA 1 LESSON 8-1
Simplify each expression.
1. 3–4
1
81
2. (–6)0
k
m–3
1
3. –2a0b–2 – 2
2
4.
5. 8000 • 40
6. 4500 • 3–2
b
8000
8-1
km3
500
Scientific Notation
ALGEBRA 1 LESSON 8-2
(For help, go to Lessons8-1.)
Simplify each expression.
1. 6 • 104
2. 7 • 10–2
3. 8.2 • 105
4. 3 • 10–3
5. 3.4 • 101
6. 5.24 • 102
7. Simplify 3  102 + 6  101 + 7  100 + 8  10–1.
8-2
Scientific Notation
ALGEBRA 1 LESSON 8-2
Solutions
1. 6 • 104 = 6 • 10,000 = 60,000
2. 7 • 10–2 = 7 • 0.01 = 0.07
3. 8.2 • 105 = 8.2 • 100,000 = 820,000
4.
5.
6.
7.
3 • 10–3 = 3 • 0.001 = 0.003
3.4 • 101 = 3.4 • 10 = 34
5.24 • 102 = 5.24 • 100 = 524
3  102 + 6  101 + 7  100 + 8  10–1
= (3  102) + (6  101) + (7  100) + (8  10–1)
= (3  100) + (6  10) + (7  1) + (8  0.1)
= 300 + 60 + 7 + 0.8
= 367.8
8-2
Scientific Notation
ALGEBRA 1 LESSON 8-2
Is each number written in scientific notation? If not, explain.
a. 0.46  104
b. 3.25  10–2
c. 13.2  106
No;
0.46 is less that 1.
yes
No;
13.2 is greater than 10.
8-2
Scientific Notation
ALGEBRA 1 LESSON 8-2
Write each number in scientific notation.
a. 234,000,000
234,000,000 = 2.34  108
Move the decimal point 8 places to the
left and use 8 as an exponent.
Drop the zeros after the 4.
b. 0.000063
0.000063 =
6.3  10–5
Move the decimal point 5 places to the
right and use –5 as an exponent.
Drop the zeros before the 6.
8-2
Scientific Notation
ALGEBRA 1 LESSON 8-2
Write each number in standard notation.
a. elephant’s mass: 8.8  104 kg
8.8  104 = 8.8000.
A positive exponent indicates a
number greater than 10. Move the
decimal point 4 places to the right.
= 88,000
b. ant’s mass: 7.3  10–5 kg
7.3  10–5 = 0.00007.3
A negative exponent indicates a
number between 0 and 1. Move the
decimal point 5 places to the left.
= 0.000073
8-2
Scientific Notation
ALGEBRA 1 LESSON 8-2
List the planets in order from least to greatest distance from
the sun.
Planet
Distance from
the Sun
Jupiter
4.84  108 mi
Earth
9.3  107 mi
Neptune
4.5  109 mi
Mercury
3.8  107 mi
Order the powers of 10. Arrange the decimals with the same power of 10
in order.
8-2
Scientific Notation
ALGEBRA 1 LESSON 8-2
(continued)
3.8  107
Mercury
9.3  107
Earth
4.84  108
Jupiter
4.5  109
Neptune
From least to greatest distance from the sun, the order of the planets is
Mercury, Earth, Jupiter, and Neptune.
8-2
Scientific Notation
ALGEBRA 1 LESSON 8-2
Order 0.0063  105, 6.03  104, 6103, and 63.1  103 from
least to greatest.
Write each number in scientific notation.
0.0063  105
6.3  102
6.03  104
63.1  103
6103
6.03  104
6.103  103
6.31  104
Order the powers of 10. Arrange the decimals with the same power of 10
in order.
6.3  102
6.103  103
6.03  104
Write the original numbers in order.
0.0063  105
6103
6.03  104
8-2
6.31  104
63.1  103
Scientific Notation
ALGEBRA 1 LESSON 8-2
Simplify. Write each answer using scientific notation.
a. 6(8  10–4) = (6 • 8)  10–4
b.
0.3(1.3 
103)
Use the Associative Property of
Multiplication.
= 48  10–4
Simplify inside the parentheses.
= 4.8  10–3
Write the product in scientific notation.
= (0.3 • 1.3)  103
Use the Associative Property of
Multiplication.
= 0.39  103
Simplify inside the parentheses.
= 3.9  102
Write the product in scientific notation.
8-2
Scientific Notation
ALGEBRA 1 LESSON 8-2
pages 402–404 Exercises
1. No; 55 > 10.
2. yes
11. 3.25  10–3
12. 8.003  106
3. No; 0.9 < 1.
13. 9.2  10–4
4. yes
14. 1.56  10–2
5. yes
15. 500
6. No; 46 > 10.
16. 0.05
7. 9.04  109
17. 2040
8. 2.0  10–2
18. 720,000
9. 9.3 
19. 0.897
106
10. 2.17 
104
22. 0.0048
23. 10–3, 10–1, 100, 101, 105
24. 6  10–10, 8  10–8, 9  10–7,
7  10–6
25. 0.52  10–3, 50.1  10–3,
4.8  10–1, 56  10–2
26. 5300  10–1, 5.3  105,
20. 1.3
21. 0.0000274
8-2
0.53  107, 530  108
27. C, A, B
28. 5.6  10–2
29. 2.4  1015
Scientific Notation
ALGEBRA 1 LESSON 8-2
30. 6.0  101
31. 3.18  10–3
32. 2.46  10–3
33. 3.4  105
34. 5400
35. 7  101
36. 1  101
37. 4.6  10–2
38. 0.0005
39. 3 
10–26
40. Answers may vary.
43.
Sample: Yes, if you
regard the 1 in
44.
1  105 as “understood”
as happens when 1 is
the coefficient of a term
like x, then 105 is in
scientific notation.
41. 48 million = 48  106.
Write 48 in scientific notation;
then add the powers of
10: 4.8  101  106 = 4.8  107.
48 millionths = 48  10–6.
45.
So 4.8  101  10–6 = 4.8  10–5.
42. about $1.65  1012
8-2
2.796  1010 instructions;
1.6776  1012 instructions
Answers may vary.
Sample: Since the
national debt is expressed
in dollars, use standard
notation, which most
people will understand,
rather than scientific
notation, which is used
mainly in science.
a. 5  1014
b. about 1.6  108 years
Scientific Notation
ALGEBRA 1 LESSON 8-2
46. about 2.61  109 people
47. a. 6.08  1010 km3
b. 1.09  1012 km3
c. 9.17  1014 km3
48.
49.
50.
51.
3.3  10–3
D
G
C
53. 4
54. 3
55. 2
3
56. 64
49
57. 1
9
58.
52. [2] (8  10–4)(1000) =
0.0008  1000 = 0.8;
the diameter is 0.8 mm
[1] minor computational error
59.
8-2
60.
Scientific Notation
ALGEBRA 1 LESSON 8-2
1. Write each number in scientific notation.
a. 0.00627
b. 3,486,000
6.27  10–3
3.486  106
2. Write each number in standard form.
a. 9.4  104
b. 2.3  10–6
94,000
0.0000023
3. Order the following numbers from least to greatest.
0.98  10–1, 1.6  103, 2.4  10–1, 11  100
0.98  10–1, 2.4  10–1, 11  100, 1.6  103
4. Simplify. Write the answer in scientific notation.
7(6.1  10–2) 4.27  10–1
8-2
Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-3
(For help, go to Lesson 1-6.)
Rewrite each expression using exponents.
1.
t•t•t•t•t•t•t
2. (6 – m)(6 – m)(6 – m)
3.
(r + 5)(r + 5)(r + 5)(r + 5)(r + 5)
4. 5 • 5 • 5 • s • s • s
Simplify.
5. –54
6.
(–5)4
7. (–5)0
8.
(–5)–4
8-3
Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-3
Solutions
1. t • t • t • t • t • t • t = t7
2. (6 – m)(6 – m)(6 – m) = (6 – m)3
3. (r + 5)(r + 5)(r + 5)(r + 5)(r + 5) = (r + 5)5
4.
5.
6.
7.
8.
5 • 5 • 5 • s • s • s = 53 • s3 = 53s3
–54 = –(5 • 5 • 5 • 5) = –(25 • 25) = –625
(–5)4 = (–5)(–5)(–5)(–5) = (25)(25) = 625
(–5)0 = 1
(–5)–4 = (– 1 )4
5
= (– 1 )(– 1 )(– 1 )(– 1 )
5
5
= ( 1 )( 1 )
25 25
= 1
625
5
5
8-3
Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-3
Rewrite each expression using each base only once.
a. 73 • 72 = 73 + 2
= 75
b. 44 • 41 • 4–2 = 44 + 1 – 2
= 43
Add exponents of powers with the
same base.
Simplify the sum of the exponents.
Think of 4 + 1 – 2 as 4 + 1 + (–2) to
add the exponents.
Simplify the sum of the exponents.
= 60
Add exponents of powers with the
same base.
Simplify the sum of the exponents.
=1
Use the definition of zero as an exponent.
c. 68 • 6–8 = 68 + (–8)
8-3
Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-3
Simplify each expression.
a.
p2 • p • p5 = p 2 + 1 + 5
= p8
Add exponents of powers with the same base.
Simplify.
b. 4x6 • 5x–4 = (4 • 5)(x 6 • x –4) Commutative Property of Multiplication
= 20(x 6+(–4))
Add exponents of powers with the
same base.
= 20x 2
Simplify.
8-3
Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-3
Simplify each expression.
a.
a 2 • b –4 • a 5 = a 2 • a 5 • b –4
Commutative Property of Multiplication
Add exponents of powers with the
same base.
= a 2 + 5 • b –4
a7
= 4
b
Simplify.
b. 2q • 3p3 • 4q4 = (2 • 3 • 4)(p 3)(q • q 4)
Commutative and Associative
Properties of Multiplication
= 24(p 3)(q 1 • q 4)
Multiply the coefficients. Write q as q 1.
= 24(p 3)(q 1 + 4)
Add exponents of powers with the
same base.
= 24p 3q 5
Simplify.
8-3
Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-3
Simplify (3  10–3)(7  10–5). Write the answer in scientific
notation.
(3 
10–3)(7

10–5) =
(3 • 7)(10–3 • 10–5)
Commutative and Associative
Properties of Multiplication
= 21  10–8
Simplify.
= 2.1  101 • 10–8
Write 21 in scientific notation.
= 2.1 
Add exponents of powers with the
same base.
101 + (– 8)
= 2.1  10–7
Simplify.
8-3
Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-3
The speed of light is 3  108 m/s. If there are1  10–3 km in
1 m, and 3.6  103 s in 1 h, find the speed of light in km/h.
Speed of light =
meters
kilometers
seconds
•
•
seconds
meters
hour
m
km
s
= (3  108) s • (1  10–3)
• (3.6  103)
m
h
= (3 • 1 • 3.6)  (108 • 10–3 • 103)
= 10.8  (108 + (– 3) + 3)
Use dimensional analysis.
Substitute.
Commutative and Associative
Properties of Multiplication
Simplify.
8-3
Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-3
(continued)
= 10.8  108
Add exponents.
= 1.08  101 • 108
Write 10.8 in scientific notation.
= 1.08  109
Add the exponents.
The speed of light is about 1.08  109 km/h.
8-3
Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-3
pages 407–410 Exercises
1. 210
12. –4.8n3
22. 6  105
5
2. 28
13. b3
23. 6  109
3. 1
14. –7
24. 4  103
4. (0.99)3
15. –45a4
25. 3.4  10–5
5. 69
3
16. y
x
26. 5.6  10–7
6. 1
17. 45x 7y 6
27. 1.5  1022
7. c5
18. 12a6c8
28. about 2.5578  1013 mi
8. 3r 5
19. x10y2
29. 1.08  1021 dollars
9. 10t –7
20. a8b
10. 56x 6
3
21. – 240m
30. about 3.84  105 km
5
r
11. 3x4
8-3
31. 9
Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-3
32. –4
44. –6x6
54. about 1.01 g
33. –3
45. 12a7
34. 11
46. x10
35. –5
47. 34 • 22
36. 5
48. 2.7  10–8
37. –4
49. 8.0  105
55. a. y1y7; y2y6; y3y5; y4y4
b. Answers may vary. Sample:
y–1y9; y–2y10; y–3y11; y–4y12
c. An infinite number; there are
an infinite number of integer
pairs with a sum of 8.
38. 0
50. 2.1  10–5
39. 2, –3
40.
6x3
+
2x2
51. 1.2  10–4
41. 4x4
52. 8.0  10–8
42. 4y5 + 8y2
53. 1.5  108
56. a. about 10–7 m
b. Longer; 1 < 4 < 7 so 1  10–7 <
4  10–7 < 7  10–7.
57. Answers may vary. Sample:
The property of multiplying powers
only applies when 2 terms have
the same base.
43. 4c4
8-3
Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-3
58. about 5.85  103 m
59. 7.65  1014
60. 4.0392  108
61. 7.039305  10–7
62. 1.7882786  10–12
63. about 6.7  1033 molecules
64. 1.428  1033 molecules
69. 8m5 + 56m3
78. 700 times
70. –8x5 + 36x4
79. D
71. 81
80. F
72. 22n+3
81. A
73. 2x+y • 3x+2
82. B
74.
1
a+b
75. (t + 3)2
65. x3
66. 1
76. 25
67. 5c3
77. a. 1.833  10–9 km3
a
68. 6a3 + 10a2
b. 1.833 m3
8-3
83. [2] 365  4.7  107
= 1715.5  107
= 1.7155  1010:
about 1.7  1010
diapers
[1] no work shown
OR answer not
written in
scientific notation
Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-3
84. [4] a. 4r 2
b. 4(5)2 = 4  25 = 100 in.2
c. 144 = 4r 2, 36 = r 2
so r = 6, d = 2r = 12, 12 in.
[3] radius found in (c) but
not the diameter
[2] only two questions
answered correctly
[1] only one question
answered correctly
89. 876,000,000
90. 0.001052
91. 910,000,000,000
92. 0.00029
93.
96. 18; 34; 46
97. –1; 7; 13
98. 4; –12; –24
85. 1.28  106
86. 3.5  10–3
95.
99. –6.8; –22.8; –34.8
94.
87. 9.0  10–5
88. 6.2  106
8-3
Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-3
Simplify each expression.
1. 34 • 35
2. 4x5 • 3x–2
39
3. (3  104)(5  102) 1.5  107
12x3
4. (7  10–4)(1.5  105) 1.05  102
5. (–2w –2)(–3w2b–2)(–5b–3) – 305
b
6. What is 2 trillion times 3 billion written in scientific notation? 6  1021
8-3
More Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-4
(For help, go to Lesson 8-3.)
Rewrite each expression using each base only once.
1. 32 • 32 • 32
2. 23 • 23 • 23 • 23
3. 57 • 57 • 57 • 57
4. 7 • 7 • 7
Simplify.
5. x3 • x3
6. a2 • a2 • a2
7. y–2 • y–2 • y–2
8. n–3 • n–3
8-4
More Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-4
Solutions
1. 32 • 32 • 32 = 3(2 + 2 + 2) = 36
2. 23 • 23 • 23 • 23 = 2(3 + 3 + 3 + 3) = 212
3. 57 • 57 • 57 • 57 = 5(7 + 7 + 7 + 7) = 528
4. 7 • 7 • 7 = 73
5. x3 • x3 = x(3 + 3) = x6
6. a2 • a2 • a2 = a(2 + 2 + 2) = a6
7. y–2 • y–2 • y–2 = y(–2 + (–2) + (–2)) = y–6 = 16
y
8. n–3 • n–3 = n(–3 + (–3)) = n–6 = 16
n
8-4
More Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-4
Simplify (a3)4.
(a3)4 = a3 • 4
= a12
Multiply exponents when raising a
power to a power.
Simplify.
8-4
More Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-4
Simplify b2(b3)–2.
b2(b3)–2 = b2 • b3 • (–2)
Multiply exponents in (b3)–2.
= b2 • b–6
Simplify.
= b2 + (–6)
Add exponents when multiplying
powers of the same base.
= b–4
Simplify.
1
= b4
Write using only positive exponents.
8-4
More Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-4
Simplify (4x3)2.
(4x3)2 = 42(x3)2
Raise each factor to the second power.
= 42x6
Multiply exponents of a power raised
to a power.
= 16x6
Simplify.
8-4
More Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-4
Simplify (4xy3)2(x3)–3.
(4xy3)2(x3)–3 = 42x2(y3)2 • (x3)–3
Raise the three factors to the second
power.
= 42 • x2 • y6 • x–9
Multiply exponents of a power raised
to a power.
= 42 • x2 • x–9 • y6
Use the Commutative Property of
Multiplication.
= 42 • x–7 • y6
Add exponents of powers with the
same base.
16y6
= x7
Simplify.
8-4
More Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-4
An object has a mass of 102 kg. The expression
102 • (3  108)2 describes the amount of resting energy in joules the
object contains. Simplify the expression.
102 • (3  108)2 = 102 • 32 • (108)2
Raise each factor within parentheses
to the second power.
= 102 • 32 • 1016
Simplify (108)2.
= 32 • 102 • 1016
Use the Commutative Property of
Multiplication.
= 32 • 102 + 16
Add exponents of powers with the
same base.
= 9  1018
Simplify.
Write in scientific notation.
8-4
More Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-4
pages 413–416 Exercises
12.
1
12g 4
24. 9  1010
1. c10
13. 36y4
25. 8  10–30
2. c10
14. 81n24
26. 8  10–9
3. n32
15.
4. q100
16. 1
5. c19
17. x16
6. d15
18. 8x5y3
7.
1
t 14
1
8y 12
19. 1
8. x 7
20. 118
9. 625y4
21. 9a6b8
10. 1024m5
22.
11. 49a2
23. 1.6  1011
c
a32
32c 26
8-4
27. 4.9  109
28. 3.6  1025
29. 6.25  10–18
30. 4.2875  10–11
31.
32.
33.
34.
35.
36.
8.57375  10–10 m3
3
–4
4
–3
0
More Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-4
37. 8
48. 9
38. –2
49. 4.3  104
39. 0
50. –256x4y5
40. –3
51. a.
41. The student who wrote
b.
x5 + x5 = 2x5 is correct;
c.
x5 times x5 is x10.
d.
24x2; 96x2
4 times
8x3; 64x3
8 times
42. 1
52. (mn)4
43. 243x3
53. (ab)5
44. b17
54. (7xyz)2
45. 30x2
46. 0.16
4
x
47. –8a9b6
57. a.
b.
c.
d.
106
109
109
1018
58. a. 223 bits
b. 230 bytes; 233 bits
59. a. about 5.15  1014 m2
b. about 3.60  1014 m2
c. about 1.37  1018 m3
60. C
61. Add exponents for
products of powers as
55. (2xy)2
in a2a4. Multiply exponents
56. Check students’ work.
for powers of powers,
as in (a2)4.
8-4
More Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-4
62. 3
73. [2] No, (x2 + 3y)2 =/ x4 + 9y2;
63. 6
for x = 2, y = 4: (x2 + 3y)2
80. (7, 26)
64. 12
= (4 + 12)2 = 162 = 256;
81. (–9, –5)
65. 3
x4 + 9y2 = 24 + 9(4)2 =
82. – 3
16 + 9 • 16 = 160
(OR equivalent explanation)
[1] appropriate conclusion
but no work shown
66. 4
67. –5
68.
x12;
x81;
no
b2
c6
69. B
74.
70. I
75. a8b3
71. C
76. 54m5n4
72. H
77. –4t5
78. (1, 8)
8-4
79. (4 2 , 1 1 )
3
4
83. 6
84. – 3
2
85. – 9
11
3
More Multiplication Properties of Exponents
ALGEBRA 1 LESSON 8-4
Simplify each expression.
1.
(x4)5
3. (5x4)3
5.
x(x5y–2)3
x20
2.
125x12
4. (1.5  105)2
(2w–2)4(3w2b–2)3
432
b6w2
x16
y6
2.25  1010
6. (3  10–5)(4  104)2
8-4
4.8  103
Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
(For help, go to Skills Handbook page 724.)
Write each fraction in simplest form.
5
1. 20
5.
6
15
9. 5xy
15x
125
2. 25
6.
8
30
10. 6y2
60
124
4
3. 100
4.
7. 10
8. 18
11. 3ac
12. 24m
12a
6mn2
35
3x
8-5
63
Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
Solutions
1.
5
= 5•1 = 1
20
5•4
4
2.
125
25 • 5
=
=5
25
25 • 1
3.
60
20 • 3
3
=
=
100
20 • 5
5
4.
124
4 • 31
=
= 31
4
4•1
5.
6
3•2 2
=
=
15
3•5 5
6.
8
2•4
4
=
=
30 2 • 15
15
7.
10
5•2
2
=
=
35
5•7
7
8.
18
9•2
2
=
=
63
9•7
7
9.
5xy
5•x•y
y
=
=
15x 5 • 3 • x 3
10. 6y2 = 3 • 2 • y2 = 2y2
3x
3•x
x
11. 3ac = 3 • a • c = c
12a
3•4•a 4
12.
8-5
24m
6•4•m
4
=
=
6mn2
6 • m • n2
n2
Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
Simplify each expression.
a.
Subtract exponents when dividing
powers with the same base.
x4
4–9
x
=
9
x
= x–5
Simplify the exponents.
1
Rewrite using positive exponents.
= x5
b.
p3 j –4
p3 – (–3)j
=
–3
6
p j
= p6 j –10
p6
= j10
–4 – 6
Subtract exponents when dividing
powers with the same base.
Simplify.
Rewrite using positive exponents.
8-5
Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
A small dog’s heart beats about 64 million beats in a year. If
there are about 530 thousand minutes in a year, what is its average
heart rate in beats per minute?
6.4  107 beats
64 million beats
=
530 thousand min
5.3  105 min
Write in scientific notation.
6.4
 107–5
Subtract exponents when dividing
powers with the same base.
6.4
 102
Simplify the exponent.
= 5.3
= 5.3
1.21  102
= 121
Divide. Round to the nearest hundredth.
Write in standard notation.
The dog’s average heart rate is about 121 beats per minute.
8-5
Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
3
y3
Simplify
3
y3
4
4
.
34
= 34
(y )
Raise the numerator and the
denominator to the fourth power.
34
= y 12
Multiply the exponent in the denominator.
81
= y 12
Simplify.
8-5
Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
a. Simplify
2
3
–3
=
–3
2
3
3
2
.
3
2
Rewrite using the reciprocal of 3 .
Raise the numerator and the
denominator to the third power.
33
= 23
27
3
= 8 or 3 8
Simplify.
8-5
Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
(continued)
4b
b. Simplify – c .
– 4b
c
–2
=
– c
4b
2
=
– c
4b
2
Rewrite using the reciprocal of – 4b .
c
Write the fraction with a negative
numerator.
(–c)2
= (4b)2
Raise the numerator and denominator
to the second power.
c2
= 16b2
Simplify.
8-5
Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
20. 1  10–12 s
t 11
pages 420–423 Exercises 12.
27m4
1. 7
13. 5  107
2. 1
14. 3  10–5
3. –3
4. 0
5. 1
4
6. 4
7. 1
c3
8. m3
2
9. s
2
10. y
21. 9
22.
15. 6  102
23.
16. 4.2  103
24.
17. 1.5  10–6
25.
18. 7  10–3
19. a. 3.86 
1011
26.
h;
2.65  108 people
b. about 1457 h
c. about 4.0 h
11. 1 2
cd
8-5
27.
28.
29.
25
1
x3
32x 5
y5
81a 4
16b 4
64
125
1
9
36
n12
8p 3
125
3
2
Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
30. 9
40. x0 simplifies to 1.
4
31. 9
4
41. The base d should
appear only once.
32. – 27
42. 9
8
33.
x8
8
1
16m12
44. 1
25
25
34. 8
n3
35. 112
c
43.
45. a6
36. 1
6
46. t
27
47. 1
n28
48. 9
4k 10
37. 53 simplifies to 125.
38. y–2 contains a negative
exponent.
39. Each term should be raised
to the 4th power and simplified.
49. 49
8-5
Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
50. a. 9.74  107 households;
5.44  1011 local calls;
9.7  1010 long-distance calls
b. about 5585 local calls
c. about 996 long-distance calls
51. a. Answers may vary.
4
Sample: c 6 can be written as
c
c4–6 or c–2; c–2 = 12 .
c
b. Check students’ work.
4
52. a 10
4b
5 5
53. a 3c
b
54. 3
4
55. 5
56.
r5
p 25q5
57. 20,736
6
58. y
81
59. 3b 5
5a
60. a. about 0.02%
b. about 99.98%
61. Answers may vary. Sample:
You can raise the numerator
and denominator to the power
and then simplify or simplify
and then raise to the power.
8-5
Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
70.
62. a. about $12,988
b. about $20,733
c. about 60%
71.
63. a. The student treated 54 as 5
b. 125
64.
3
5
65. m
n
5
7
66. d 3
67. 1010
68. 3x 3
69.
2y
2
13m
2
5
5
4
.
7m
5n
5c
6
2
3
72. a. about 1.2  106 s
b. about 13.9 days
73. a-b. Check students’ work.
c. No, the power may remain
the same or be one less.
74. def. of neg. exponent
75. dividing powers with the same base,
def. of neg. exponent
76. raising a quotient to a power
77. mult. powers with the same base
78. raising a power to a power, dividing
powers with the same base, def. of
neg. exponents
8-5
Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
79. n2
80.
b. The closer the ratio is to 1,
the more circular the orbit.
c. Pluto, Venus
n4x
81. x4
84. B
82. 1
n2
85. G
83. a.
86. D
87. B
88. C
89. B
8-5
Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
90. [4] (7.43  108) ÷ (2.5  104) = 2.972  104,
29,720 h; 29,720 hours •
1
1 day • 1 year
24 hours 365 days
3.4 years
about 3.4 years (OR equivalent explanation)
[3] appropriate methods but one computational error
[2] error in conversion factor used or a missing conversion factor
[1] correct number of hours and years but no work shown
97. n4
91. 27y6
100. (–4, –7)
8
m 21
t 20
93. 8
r
6
94. 2s
27
95. 1
8c 2
92.
98. 1
99. (0, 0)
96. 9r6
8-5
Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
101. (3, 5)
103.
102. no solution
8-5
Division Properties of Exponents
ALGEBRA 1 LESSON 8-5
Simplify each expression.
1.
a8
a–2
4.
1.6  103
2.
a10
4  10–2
4  104
5.
w3
w7
24
5
8-5
1
w4
2
256
6
or
10
25
25
4
–2
3. (3a) (2a )
6a2
6.
4x 3
3x 2
27
–3
27
64x3
Geometric Sequences
ALGEBRA 1 LESSON 8-6
(For help, go to Lesson 5-6.)
Find the common difference of each sequence.
1. 1, 3, 5, 7, ...
2. 19, 17, 15, 13, ...
3. 1.3, 0.1, –1.1, –2.3, ...
4. 18, 21.5, 25, 28.5, ...
Use inductive reasoning to find the next two numbers in
each pattern.
5. 2, 4, 8, 16, ...
6. 4, 12, 36, ...
7. 0.2, 0.4, 0.8, 1.6, ...
8. 200, 100, 50, 25, ...
8-6
Geometric Sequences
ALGEBRA 1 LESSON 8-6
Solutions
1. 1, 3, 5, 7, ...
7 – 5 = 2, 5 – 3 = 2, 3 – 1 = 2
Common difference: 2
2.
19, 17, 15, 13, ...
13 – 15 = –2, 15 – 17 = –2,
17 – 19 = –2
Common difference: –2
3. 1.3, 0.1, –1.1, –2.3, ...
–2.3 – (–1.1) = –1.2, –1.1
– 0.1 = –1.2, 0.1 – 1.3 = –1.2
Common difference: –1.2
4.
18, 21.5, 25, 28.5, ...
28.5 – 25 = 3.5, 25 – 21.5 = 3.5,
21.5 – 18 = 3.5
Common difference: 3.5
8-6
Geometric Sequences
ALGEBRA 1 LESSON 8-6
Solutions (continued)
5. 2, 4, 8, 16, ...
2(2) = 4, 4(2) = 8, 8(2) = 16,
16(2) = 32, 32(2) = 64
Next two numbers: 32, 64
6. 4, 12, 36, ...
4(3) = 12, 12(3) = 36,
36(3) = 108, 108(3) = 324
Next two numbers: 108, 324
7. 0.2, 0.4, 0.8, 1.6, ...
(0.2)2 = 0.4, 0.4(2) = 0.8, 0.8(2) =
1.6, 1.6(2) = 3.2, 3.2(2) = 6.4
Next two numbers: 3.2, 6.4
8. 200, 100, 50, 25, ...
200  2 = 100, 100  2 = 50,
50  2 = 25, 25  2 = 12.5,
12.5  2 = 6.25
8-6
Geometric Sequences
ALGEBRA 1 LESSON 8-6
Find the common ratio of each sequence.
a. 3, –15, 75, –375, . . .
–15
3
(–5)
–375
75
(–5)
(–5)
The common ratio is –5.
3
3
3
b. 3, 2 , 4 , 8 , ...
3
2
3

1
2
3
4

3
8
1
2

1
2
1
The common ratio is 2 .
8-6
Geometric Sequences
ALGEBRA 1 LESSON 8-6
Find the next three terms of the sequence 5, –10, 20, –40, . . .
5
–10
(–2)
20
(–2)
–40
(–2)
The common ratio is –2.
The next three terms are –40(–2) = 80, 80(–2) = –160, and –160(–2) = 320.
8-6
Geometric Sequences
ALGEBRA 1 LESSON 8-6
Determine whether each sequence is arithmetic
or geometric.
a. 162, 54, 18, 6, . . .
62
54
1
3
18
1
3
6
1
3
The sequence has a common ratio.
The sequence is geometric.
8-6
Geometric Sequences
ALGEBRA 1 LESSON 8-6
(continued)
b. 98, 101, 104, 107, . . .
98
101
+3
104
+3
107
+3
The sequence has a common difference.
The sequence is arithmetic.
8-6
Geometric Sequences
ALGEBRA 1 LESSON 8-6
Find the first, fifth, and tenth terms of the sequence that has
the rule A(n) = –3(2)n – 1.
first term: A(1) = –3(2)1 – 1 = –3(2)0 = –3(1) = –3
fifth term: A(5) = –3(2)5 – 1 = –3(2)4 = –3(16) = –48
tenth term: A(10) = –3(2)10 – 1 = –3(2)9 = –3(512) = –1536
8-6
Geometric Sequences
ALGEBRA 1 LESSON 8-6
Suppose you drop a tennis ball from a height of 2 meters.
On each bounce, the ball reaches a height that is 75% of its previous
height. Write a rule for the height the ball reaches on each bounce. In
centimeters, what height will the ball reach on its third bounce?
The first term is 2 meters, which is 200 cm.
Draw a diagram to help understand the problem.
8-6
Geometric Sequences
ALGEBRA 1 LESSON 8-6
(continued)
The ball drops from an initial height, for which there is no bounce. The
initial height is 200 cm, when n = 1. The third bounce is n = 4. The
common ratio is 75%, or 0.75.
A rule for the sequence is A(n) = 200 • 0.75n – 1.
A(n) = 200 • 0.75n – 1
Use the sequence to find the height of
the third bounce.
A(4) = 200 • 0.754 – 1
Substitute 4 for n to find the height of
the third bounce.
= 200 • 0.753
Simplify exponents.
= 200 • 0.421875
Evaluate powers.
= 84.375
Simplify.
The height of the third bounce is 84.375 cm.
8-6
Geometric Sequences
ALGEBRA 1 LESSON 8-6
12. –48, 96, –192
24. –1.1; 70.4; 18,022.4
1. 4
13. geometric
25. A(n) = 6 • 0.5n–1; 0.375
2. 4
14. arithmetic
26. A(n) = –6 • 2n–1; –3072
3. 0.1
15. geometric
27. A(n) = 7 • (1.1)n–1; 9.317
4. 2.50
16. arithmetic
28. A(n) = 1 • (-4)n–1; 4096
5. –0.25
17. arithmetic
6. 2
18. geometric
29. a. A(n) = 100 • (0.64)n–1
b. about 10.74 cm
7. 40, 80, 160
19. 5; 135; 10,935
8. 48, 96, 192
20. –5; –135; –10,935
pages 427–429 Exercises
9. 20.25, 30.375, 45.5625 21. 5; –135; –10,935
10. –0.5, 0.25, –0.125
22. 0.5; 13.5, 1093.5
11. 0.36, 0.072, 0.0144
23. –2; –250; –156,250
8-6
30. 2 2 , 8 , 8 ; A(n) =
3 9 27
n–1
216 • 1
3
31. 1, 0.2, 0.04; A(n) =
625 • (0.2)n–1
Geometric Sequences
ALGEBRA 1 LESSON 8-6
32. 656.1, 5904.9, 53,144.1; A(n) = 0.1 • 9n–1 41. a. A(n) = 36 • (0.9)n–1
b. 6; n = 1 corresponds to the
33. 1, –0.5, 0.25; A(n) = 16 • (–0.5)n–1
first swing, because A(1) = 36.
34. Check students’ work.
c. 21.3 cm
35. If all consecutive terms have a common
difference, the sequence is arithmetic.
If all consecutive terms have a common
ratio, the sequence is geometric.
42. a. 1 , 1 , 1 , 1
2 4
b.
2–1,
8
2–2,
16
2–3, 2–4
c. r = 2–n
d. 2–10 or 110
36. a. 3 folds: 8; 4 folds: 16; 5 folds: 32
b. A(n) = 1 • 2n
c. 1024 rectangles
2
43. No; if a term were 0, then all terms
would be 0 because you multiply
a term to get the next term.
37. arithmetic; 3, 1, –1
38. neither; –3, –8, –14
39. geometric; 1.125, 0.5625, 0.28125
40. arithmetic; 20, 22, 24
8-6
Geometric Sequences
ALGEBRA 1 LESSON 8-6
49. 6  10–2; 2.592  102,
44. a. 1, 3 , 9 , 27
4 16 64
n–1
b. r = 1 • 3
4
243
c.
1024
d. 0, 1 , 7 , 37
4 16 64
e. r =
1.5552  101, 9.3312  10–1
50. 5th
51. D
52. H
1–1• 3
n–1
4
f. 14,197
16,384
45. x; x5, x6, x7
46. 3x; 27x4, 81x5, 243x6
47. xy2; x5y9, x6y11, x7y13
48. ab; 2a4b2, 2a5b3, 2a6b4
53. B
54. [2] Day 1 350
Day 2 700
Day 3 1400
Day 4 2800
Day 5 5600
(OR equivalent explanation)
[1] answer with no work shown
8-6
Geometric Sequences
ALGEBRA 1 LESSON 8-6
55. a4
66. y = – 2 x
56. 16
57. 101 20
x z
58. 1
59.
60.
61.
62.
67. y =
68. y =
69. y =
81
64
1
m14
1
p15
d5
c2
70. y =
71. y =
72. y =
5
– 7x
6
5x
3
7x
4
– 1x
4
5x
9
–1x
2
63. 2.467  10–3
64. 3.26  1011 gal
65. y = 8 x
3
8-6
Geometric Sequences
ALGEBRA 1 LESSON 8-6
1. Find the common ratio of the geometric sequence –3, 6, –12, 24, . . .
–2
2. Find the next three terms of the sequence 243, 81, 27, 9, . . .
1
3, 1,
3
3. Determine whether each sequence is arithmetic or geometric.
a. 37, 34, 31, 28, . . . arithmetic
b. 8, –4, 2, –1, . . .
geometric
4. Find the first, fifth, and ninth terms of the sequence that has the rule
A(n) = 4(5)n–1. 4, 2500, 1,562,500
5. Suppose you enlarge a photograph that is 4 in. wide and 6 in. long so
that its dimensions are 20% larger than its original size. Write a rule for
the length of the copies. What will be the length if you enlarge the
photograph five times? (Hint: The common ratio is not just 0.2. You
must add 20% to 100%.)
A(n) = 6(1.2)n-1; about 14.9 in.
8-6
Exponential Functions
ALGEBRA 1 LESSON 8-7
(For help, go to Lesson 5-6.)
Graph each function.
1. y = 3x
2. y = 4x
3. y = –2x
Simplify each expression.
4. 32
5. 5–3
6. 2 • 34
7. 2 • 3–2
8. 3 • 2–1
9. 10 • 32
8-7
Exponential Functions
ALGEBRA 1 LESSON 8-7
Solutions
1. y = 3x
3. y = –2x
2. y = 4x
4. 32 = 3 • 3 = 9
5. 5–3 = 13 =
5
6. 2 • 34 = 2 • (3 • 3 • 3 • 3) = 2 • 81 = 162
7. 2 • 3–2 = 2 • 12 = 2 • 1 = 2
3
9 9
8. 3 • 2-1 = 3 • 1 = 3 • 1 = 3 or 1 1
21
2 2
2
9. 10 • 32 = 10 • (3 • 3) = 10 • 9 = 90
8-7
1
= 1
5 • 5 • 5 125
Exponential Functions
ALGEBRA 1 LESSON 8-7
Evaluate each exponential function.
a. y = 3x for x = 2, 3, 4
x
2
3
4
y = 3x
32 = 9
33 = 27
34 = 81
y
9
27
81
b. p(q) = 3 • 4q for the domain {–2, 3}
q
p(q) = 3 • 4q
–2
3 • 4–2 = 3 • 16 = 16
3
3 • 43 = 3 • 64 = 192 192
1
p(q)
3
3
16
8-7
Exponential Functions
ALGEBRA 1 LESSON 8-7
Suppose two mice live in a barn. If the number of mice
quadruples every 3 months, how many mice will be in the barn after
2 years?
ƒ(x) = 2 • 4x
ƒ(x) = 2 • 48
In two years, there are 8 three-month
time periods.
ƒ(x) = 2 • 65,536
Simplify powers.
ƒ(x) = 131,072
Simplify.
8-7
Exponential Functions
ALGEBRA 1 LESSON 8-7
Graph y = 2 • 3x.
x
y = 2 • 3x
–2
2 • 3–2 = 3 2 =
–1
2 • 3–1 =
(x, y)
2
2
31
=
2
9
(–2, 29 )
2
3
(–1, 3 )
2
0
2 • 30 = 2 • 1 = 2
(0, 2)
1
2 • 31 = 2 • 3 = 6
(1, 6)
2
2 • 32 = 2 • 9 = 18 (2, 18)
8-7
Exponential Functions
ALGEBRA 1 LESSON 8-7
The function ƒ(x) = 1.25x models the increase in size of an
image being copied over and over at 125% on a photocopier. Graph
the function.
x
ƒ(x) = 1.25x
(x, ƒ(x))
1
1.251 = 1.25
2
1.252 = 1.5625
1.6
(2, 1.6)
3
1.253 = 1.9531
2.0
(3, 2.0)
4
1.254 = 2.4414
2.4
(4, 2.4)
5
1.255 = 3.0518
3.1
(5, 3.1)
1.3
(1, 1.3)
8-7
Exponential Functions
ALGEBRA 1 LESSON 8-7
pages 432–435 Exercises
12. A
1. 216
13. C
2. 2
14. B
9
3. 2.5
4. 32
20.
15. B
16. D
21.
17. C
5. 4.5
18. A
6. 115.2
19.
22.
7. 1600
8. 0.576
9. $160,000; $320,000
10. $2000; $4000
11. $16,000, $32,000
8-7
Exponential Functions
ALGEBRA 1 LESSON 8-7
23.
27. 100, 10, 1, 0.1, 0.01, 0.001; decrease
28. 0.3125, 1.25, 5, 20, 80, 320; increase
29. 4, 2, 1, 0.5, 0.25, 0.125; decrease
30. 9 , 3 , 1, 2 , 4 , 8 ; decrease
4 2
24.
3
9 27
31. 0.04, 0.4, 4, 40, 400, 4000; increase
32. 1111.1, 333.3, 100, 30, 9, 2.7; decrease
33. a. Answers may vary.
Sample: y = 2(3)x
b. Answers may vary.
Sample: y = 2(0.3)x
25. 0.04, 0.2, 1, 5, 25, 125; increase
26. 0.16, 0.4, 1, 2.5, 6.25, 15.625; increase
8-7
Exponential Functions
ALGEBRA 1 LESSON 8-7
34. a.
b. y = 75 • 2x, where x is the number of 20-min time periods
8-7
Exponential Functions
ALGEBRA 1 LESSON 8-7
35. a.
b. (0, 1)
c. No; there is no value
of x for which y = 0.
d. If the base is > 1, the
graph gets steeper as
the base increases. If the
base is < 1, the graph gets
steeper as the base
decreases.
36. a. 1,000,000,000 plants
b. 1,000,000,000,000 plants
x
y
1
–2
2
4
3
–8
4
16
5 –32
b. Every other value is negative.
The absolute value of each term
is double the previous term.
c. No; in y = a • bx, b > 0. –2 < 0,
so it is not exponential.
38. y = x5
37. a.
39. ƒ(t) = 200 • t 2
40. y = 3x
41. ƒ(x) = 100x2
42. {500}; b = 1 produces a linear graph.
8-7
Exponential Functions
ALGEBRA 1 LESSON 8-7
43. a.
46. 6
55. [2]
47. 3
48. 4
49. 5
The graphs intersect
between x = 1 and x = 2
(OR equivalent explanation).
50. a. 4
b. Between x = 1 and x = 3,
the graph of y = x2 rises
faster than the graph of
y = 2x. The graphs
intersect at x = 2.
c. The graph of y = 6x is
steeper than y = x2
and y = 2x.
44. 2
b. 3
c. y = 4 • 3x
[1] answer with no work shown
d. 4 , 324
9
51. B
52. G
53. A
54. H
45. –3
8-7
56. 5; 1250, 6250, 31,250
57. –3; 567, –1701, 5103
58. 2; –3.2, –6.4, –12.8
59. – 1 ; 1 , – 1 , 1
3 3
9 27
Exponential Functions
ALGEBRA 1 LESSON 8-7
60. 0.1; 0.045, 0.0045, 0.00045
61. 0.25; 28, 7, 1.75
62. y = 5x
63. y = 3x + 1
64. y = –2x + 8
65. y = 0.4x – 3.8
8-7
Exponential Functions
ALGEBRA 1 LESSON 8-7
1. Evaluate each function rule for the given value.
a. y = 0.5x for x = 3 0.125
b. ƒ(x) = 4 • 3x for x = –2
4
9
2. Suppose an investment of $5000 doubles every 12 years.
a. How much is the investment worth after 24 years?
$20,000
b. After 48 years?
3. Graph y = 0.5 • 3x.
$80,000
4. Graph y = –0.5 • 3x.
8-7
Exponential Growth and Decay
ALGEBRA 1 LESSON 8-8
(For help, go to Lesson 4-3.)
Use the formula I = prt to find the interest for principal p, interest rate r,
and time t in years.
1. principal: $1000; interest rate: 5%; time: 2 years
2. principal: $360; interest rate: 6%; time: 3 years
3. principal: $2500; interest rate: 4.5%; time: 2 years
4. principal: $1680; interest rate: 5.25%; time: 4 years
5. principal: $1350; interest rate: 4.8%; time: 5 years
8-8
Exponential Growth and Decay
ALGEBRA 1 LESSON 8-8
Solutions
1. l = prt = ($1000)(0.05)(2) = $50(2) = $100
2. l = prt = ($360)(0.06)(3) = $21.60(3) = $64.80
3. l = prt = ($2500)(0.045)(2) = $112.50(2) = $225
4. l = prt = ($1680)(0.0525)(4) = $88.20(4) = $352.80
5. l = prt = ($1350)(0.048)(5) = $64.80(5) = $324
8-8
Exponential Growth and Decay
ALGEBRA 1 LESSON 8-8
In 1998, a certain town had a population of about 13,000
people. Since 1998, the population has increased about 1.4% a year.
a. Write an equation to model the population increase.
Relate: y = a • bx
Define:
Write:
Use an exponential function.
Let x = the number of years since 1998.
Let y = the population of the town at various times.
Let a = the initial population in 1998, 13,000 people.
Let b = the growth factor, which is
100% + 1.4% = 101.4% = 1.014.
y = 13,000 • 1.014x
8-8
Exponential Growth and Decay
ALGEBRA 1 LESSON 8-8
(continued)
b.
Use your equation to find the approximate population in 2006.
y = 13,000 • 1.014x
y = 13,000 • 1.0148
14,529
2006 is 8 years after 1998, so
substitute 8 for x.
Use a calculator. Round to the nearest
whole number.
The approximate population of the town in 2006 is 14,529 people.
8-8
Exponential Growth and Decay
ALGEBRA 1 LESSON 8-8
Suppose you deposit $1000 in a college fund that pays 7.2%
interest compounded annually. Find the account balance after 5 years.
Relate: y = a • bx
Define:
Write:
Use an exponential function.
Let x = the number of interest periods.
Let y = the balance.
Let a = the initial deposit, $1000
Let b = 100% + 7.2% = 107.2% = 1.072.
y = 1000 • 1.072x
= 1000 • 1.0725
1415.71
Once a year for 5 years is 5 interest
periods. Substitute 5 for x.
Use a calculator. Round to the
nearest cent.
The balance after 5 years will be $1415.71.
8-8
Exponential Growth and Decay
ALGEBRA 1 LESSON 8-8
Suppose the account in the above problem paid interest
compounded quarterly instead of annually. Find the account balance
after 5 years.
Relate: y = a • bx
Define:
Use an exponential function.
Let x = the number of interest periods.
Let y = the balance.
Let a = the initial deposit, $1000
Let b = 100% + 7.2%
There are 4 interest periods in
4
1 year, so divide the interest into
4 parts.
= 1 + 0.018 = 1.018
8-8
Exponential Growth and Decay
ALGEBRA 1 LESSON 8-8
(continued)
Write:
y = 1000 • 1.018x
= 1000 • 1.01820
1428.75
Four interest periods a year for
5 years is 20 interest periods.
Substitute 20 for x.
Use a calculator.
Round to the nearest cent.
The balance after 5 years will be $1428.75.
8-8
Exponential Growth and Decay
ALGEBRA 1 LESSON 8-8
Technetium-99 has a half-life of 6 hours. Suppose a lab has
80 mg of technetium-99. How much technetium-99 is left after
24 hours?
In 24 hours there are four 6-hour half lives.
After one half-life, there are 40 mg.
After two half-lives, there are 20 mg.
After three half-lives, there are 10 mg.
After four half-lives, there are 5 mg.
8-8
Exponential Growth and Decay
ALGEBRA 1 LESSON 8-8
Suppose the population of a certain endangered species
has decreased 2.4% each year. Suppose there were 60 of these
animals in a given area in 1999.
a. Write an equation to model the number of animals in this species that
remain alive in that area.
Relate: y = a • bx
Define:
Write:
Use an exponential function.
Let x = the number of years since 1999
Let y = the number of animals that remain
Let a = 60, the initial population in 1999
Let b = the decay factor, which is 100% - 2.4 % = 97.6% = 0.976
y = 60 • 0.976x
8-8
Exponential Growth and Decay
ALGEBRA 1 LESSON 8-8
(continued)
b. Use your equation to find the approximate number of animals
remaining in 2005.
y = 60 • 0.976x
y = 60 • 0.9766
52
2005 is 6 years after 1999, so
substitute 6 for x.
Use a calculator. Round to the
nearest whole number.
The approximate number of animals of this endangered species remaining
in the area in 2005 is 52.
8-8
Exponential Growth and Decay
ALGEBRA 1 LESSON 8-8
pages 441–444 Exercises
10. 1.005
1. 20; 2
11. 0.75%, 0.25%
21. a. 3 half-lives
b. 3.125 mCi
2. 200; 1.0875
12. 1%; 0.3%
22. 0.5
3. 10,000; 1.01
13. 1.125%; 0.375%
23. 0.1
4. 1; 1.5
14. 1.9%; 0.63%
24. 2
5. a. 50,000
b. 0.03; 1.03
c. 1.03
d. 50,000; 1.03; x
e. about 104,689 people
6. 1.04
15. 1.5625%; 0.52083% 25. 0.9
26. exp. growth
16. $5352.90
7. 1.05
8. 1.037
9. 1.0875
3
17. $16,661.35
27. exp. decay
18. $634.87
28. exp. growth
19. $28,338.18
29. exp. decay
20. a. 4 half-lives
b. 2.5 mCi
30. a. $22,000; 0.8
b. y = 22,000 • (0.8)x
c. $5767.17
8-8
Exponential Growth and Decay
ALGEBRA 1 LESSON 8-8
31. y = 130,000 • (1.01)x; about 142,179 people
40. linear function
32. y = 3,000,000 • (0.985)x;
about 2,579,191 people
33. y = 2400 • (1.07)x; $4721.16
41. exponential function
34. y = 2400 • (1.00583333)x; $4823.19
35. a. y = 584 • (1.065)x; $2057.81
b. Check students’ work.
36. Linear function; it is a straight line.
37. Neither; it is not just one straight line.
38. Exponential function; it is a curve with
y-values that increase as x-values increase.
39. Neither; it decreases and then
increases, unlike an exponential function.
8-8
42. linear function
Exponential Growth and Decay
ALGEBRA 1 LESSON 8-8
43. Answers may vary.
Sample: $600; even after
10 years, there is more
money in the account with
an initial deposit of $600
($977.34) than there is in
the account with an initial
deposit of $500 ($907.01).
44. 6 half-lives
48. 94%
49. 88%
50. 96.5%
51. 46.1%
52. a. 2 years
b. 4 years
45. 4 half-lives
46. a.
b.
c.
about 4 h
1
4
about 3.7 mg using the function and
53. a.
b.
c.
d.
e.
f.
$220.00
$3.96
$223.96
$193.96
9 months
$18.07
15 mg  4 = 3.75 mg using the prediction
54. Check students’ work.
x
47. a. y = 6,284,000 • (1.01)
55. 2003
b. 7,667,674 people
1
8-8
Exponential Growth and Decay
ALGEBRA 1 LESSON 8-8
56. C
61.
57. H
58. A
59. [2] Using $1000 for deposit, quarterly:
1000(1 + 0.05
)20 1282.04; annually:
4
1000(1 + 0.055)5 1306.96. The
account paying 5.5% will be greater.
(OR equivalent explanation)
[1] correct approach with minor
computational error
60.
62.
63. 7.28  1011 gal
8-8
Exponential Growth and Decay
ALGEBRA 1 LESSON 8-8
1. Identify the original amount a and the growth factor b in the exponential
function y = 10 • 1.036x.
a = 10, b = 1.036
2. A population of 24,500 people has been increasing at a rate of 1.8% a
year. What will be the population in 15 years if it continues at that rate?
about 32,017 people
3. Write an exponential function to represent $2000 principal earning
5.6% interest compounded annually.
y = 2000 • 1.056x
4. Find the account balance on $3000 principal earning 6.4% interest
compounded quarterly for 7 years.
about $4678.91
5. The half-life of a certain substance is 4 days. If you have 100 mg of the
substance, how much of it will remain after 12 days?
12.5 mg
6. The value of a $1200 computer decreases 27% annually. What will be
the value of the computer after 3 years?
about $466.82
8-8
Exponents and Exponential Functions
ALGEBRA 1 CHAPTER 8
3
1. r12
t
4
2. m12
a
5
3. m8
t
4. c 2v 9
5.
6.
7.
8.
h 2d 2
k3
9j 2
y5
p 35
w 14
h 26
32y 9
9. 1.728
10. –27
q
11. C
12. 4.4909 
107
3
20. a. A(n) = 12 •
5
votes
13. 4.5  105 reptiles
b. 2.592 ft
14. No; 76 > 10.
21. –192
15. yes
22. 15, 75, 375
16. no; two powers of 10
23. 2, 8, 32
17. No; 32.5 > 10.
18. a. about 6.7068  108 mi
b. about 1.5 h
24. 3.8, 3.61, 3.4295
25. 3.75, 2.8125, 2.109375
26.
19. a. – 1
2
b. –2, 1, – 1
2
c. A(n) = –32 – 1
d. – 1
8
n–1
n–1
2
8-A
Exponents and Exponential Functions
ALGEBRA 1 CHAPTER 8
27.
30. growth for b > 1, decay for 0 < b < 1
31. $1082.86; $1220.19
28.
32. a.
b.
c.
d.
growth; 1.07
about 7.1 kw-h
about 0.66 kw-h
Check students’ work.
33. a. 0.85; the car’s value depreciates
15% annually.
b. $17,000
c. $10,440
29. Answers may vary. Sample:
A computer loses 20% of its
value each year. How much
will a $3500 computer be
worth in 3 years? $1792
34. a. 8%
b. $10,000; $12,597.12
c. Check students’ work.
8-A