Chapter 6
Chapter 6 Maintaining Mathematical Proficiency (p. 289)
14
1. 12 — − 33 + 15 − 92 = 12(7) − 33 + 15 − 92
2
= 12(7) − 27 + 15 − 81
( )
⋅
8 ÷ 22 + 20
⋅
an = a1 + (n − 1)d
an = 22 + (n − 1)(−7)
= 84 − 27 + 15 − 81
an = 22 + n(−7) − 1(−7)
= 57 + 15 − 81
an = 22 − 7n + 7
= 72 − 81
an = −7n + 29
= −9
2. 52
10. The first term is 22, and the common difference is −7.
3 − 4 = 25
11. yes; no; The product of two perfect squares can be
⋅
8 ÷ 4 + 20
⋅
⋅3−4
= 200 ÷ 4 + 20
= 50 + 20
⋅
represented by m2n2 = (mm)(nn) = (mn)(mn) = (mn)2. If m
and n are integers, their product is also an integer. So, (mn)2
is an integer. There are many counterexamples illustrating
that the quotient of two perfect squares does not have to be
a perfect square, such as 9 ÷ 4.
3−4
3−4
= 50 + 60 − 4
= 110 − 4
Chapter 6 Mathematical Practices (p. 290)
= 106
1.
3. −7 + 16 ÷ 24 + (10 − 42) = −7 + 16 ÷ 24 + (10 − 16)
= −7 + 16 ÷
24
+ (−6)
Year
1
= −12
—
4. Because 82 = 64, √ 64 = √ 82 = 8.
—
5. −√ 4 represents the negative square root. Because
—
—
22
= 4,
−√4 = −√ 22 = −2.
—
6. −√ 25 represents the negative square root. Because 52 = 25,
—
—
−√25 = − √52 = −5.
8
9
10
2
3
4
5
6
7
1
2
4
8
16
32
64 128 256 512
10
6.1 Explorations (p. 291)
⋅ 2) (2 ⋅ 2 ⋅ 2)
⋅2⋅2⋅2⋅2
1. a. i. (22)(23) = (2
=2
= 25
ii.
(41)(45)
⋅ 4 ⋅ 4 ⋅ 4 ⋅ 4)
=4⋅4⋅4⋅4⋅4⋅4
= (4)(4
= 46
an = 12 + (n − 1)2
iii. (53)(55) = (5
=5
an = 12 + n(2) − 1(2)
⋅ 5 ⋅ 5)(5 ⋅ 5 ⋅ 5 ⋅ 5 ⋅ 5)
⋅5⋅5⋅5⋅5⋅5⋅5⋅5
= 58
an = 12 + 2n − 2
9. The first term is 6, and the common difference is −3.
9
So, the sum of the numbers in the tenth row is 512.
an = a1 + (n − 1)d
an = 2n + 10
8
⤻
⤻⤻⤻⤻⤻⤻⤻⤻
×2 ×2 ×2 ×2 ×2 ×2 ×2 ×2 ×2
8. The first term is 12, and the common difference is 2.
an = 6 + n(−3) − 1(−3)
7
1
—
Because 112 = 121, ±√121 = ±√112 = ±11.
an = 6 + (n − 1)(−3)
6
Sum
—
an = a1 + (n − 1)d
5
2. Row
7. ±√ 121 represents the positive or negative square root.
—
4
So, the population in the tenth year is 3435 rabbits.
= −6 + (−6)
—
3
⤻⤻⤻⤻⤻⤻⤻
⤻⤻
×1.6 ×1.6 ×1.6 ×1.6 ×1.6 ×1.6 ×1.6 ×1.6 ×1.6
= −7 + 16 ÷ 16 + (−6)
= −7 + 1 + (−6)
2
Population 50 80 128 205 328 524 839 1342 2147 3435
iv.
(x2)(x6)
⋅ x)( x ⋅ x ⋅ x ⋅ x ⋅ x ⋅ x)
=x⋅x⋅x⋅x⋅x⋅x⋅x⋅x
= (x
= x8
In each example, the exponent of the product is the sum of the
exponents of the factors. So, a general rule is aman = am+n.
an = 6 − 3n + 3
an = −3n + 9
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All rights reserved.
Algebra 1
Worked-Out Solutions
305
Chapter 6
⋅ ⋅
⋅
⋅ ⋅ ⋅ ⋅
⋅ ⋅
⋅
⋅ ⋅ ⋅ ⋅ ⋅
⋅ ⋅
⋅ ⋅
⋅ ⋅ ⋅
⋅ ⋅ ⋅
43 4 4 4 4
4 4
1
4
25 2 2 2 2 2 2 2 2
ii. —2 = —— = — = 23
2 2
1
2
x x x
x6 x x x x x x
iii. —3 = —— = — = x3
x x x
1
x
34 3 3 3 3 1
0
iv. —4 = —— = — = 3
3 3 3 3 1
3
So, to find a power of a product, find the power of each
factor and multiply, or (ab)m = ambm.
b. i. —2 = — = — = 41
( 32 ) = ( 23 )( 32 ) = 32 ⋅⋅ 23 = 23
( 43 ) = ( 43 )( 43 )( 43 ) = 34 ⋅⋅ 43 ⋅⋅ 43 = 43
— —
3
ii. —
2
—
— — —
—2
—
3
—3
( 2x ) = ( 2x )( 2x )( 2x ) = x2 ⋅⋅ x2 ⋅⋅ x2 = 2x
a
a a a a
a a a a
a
=
iv. ( ) = ( )( )( )( ) = ⋅ ⋅ ⋅
b b b b
b⋅b⋅b⋅b
b
b
—
3
— — —
4
— — — —
—
3
—3
In each example, the exponent of the quotient is the
difference of the exponents of the powers that are being
am
divided. So, a general rule is —n = am−n.
a
iii.
c. i. (22)4 = (22)(22)(22)(22)
So, to find the power of a quotient, find the power of the
numerator and the power of the denominator and divide,
a m am
.
or — = —
b
bm
⋅ 2)(2 ⋅ 2)(2 ⋅ 2)(2 ⋅ 2)
=2⋅2⋅2⋅2⋅2⋅2⋅2⋅2
—
= (2
ii.
(73)2
=
Then express the pattern using variables.
⋅ 7 ⋅ 7) (7 ⋅ 7 ⋅ 7)
=7⋅7⋅7⋅7⋅7⋅7
3. Because each side of the large cube has the same length
as 9 small cubes, an expression for the number of small
cubes in the large cube is 9 9 9 = 93.
⋅ ⋅
76
iii. (y3)3 = ( y3)( y3)( y3)
⋅ ⋅ ⋅ ⋅ ⋅ ⋅
⋅y⋅y⋅y⋅y⋅y⋅y⋅y⋅y
= (y
=y
y
y)(y
y
y)(y
y
y)
= y9
iv. (x4)2 = (x4)(x4)
⋅ ⋅ ⋅ ⋅ ⋅ ⋅
⋅x⋅x⋅x⋅x⋅x⋅x⋅x
= (x
=x
x
x
x)(x
x
x
x)
= x8
6.1 Monitoring Progress (pp. 292–295)
1. (−9)0 = 1
1
3
⋅ 5)
−50
22 = −1 4 = −4
2
1
1
3−2x−5 —
4. —
=
=—
1 32 x5 9x5
y0
3. —
= −1
−2
5. 104
⋅ 5)(2 ⋅ 5)
⋅5⋅2⋅5
=2⋅2⋅5⋅5
=2 ⋅5
ii. (5 ⋅ 4) = (5 ⋅ 4)(5 ⋅ 4)(5 ⋅ 4)
=5⋅4⋅5⋅4⋅5⋅4
=5⋅5⋅5⋅4⋅4⋅4
=5 ⋅4
2
1
27
2. 3−3 = —3 = —
In each example, the exponent of the single power is the
product of the other two exponents. So, a general rule is
(am)n = amn.
d. i. (2
⋅
= 10−2
1
= —2
10
1
=—
100
6. x9
2
⋅x
−9
⋅a⋅6⋅a
=6⋅6⋅a⋅a
=6 ⋅a
=6
2
iv. (3x)2 = (3x)(3x)
⋅x⋅3⋅x
=3⋅3⋅x⋅x
=3 ⋅x
=3
2
2
Algebra 1
Worked-Out Solutions
= x0
−58
−5
7. —4 = 1
3
iii. (6a)2 = (6a)(6a)
= x9+(−9)
=1
3
3
⋅
⋅ ⋅
=2
2
⋅
10−6 = 104+(−6)
= (2
2
4
—4
2. Sample answer: Try several examples to find a pattern.
(73)(73)
= (7
=
——
()
= 28
306
2
—
e. i.
⋅5
8−4
= 54
= 625
y6
y
8. —7 = y6−7
= y−1
1
=—
y
9. (6−2)−1 = 6(−2)(−1)
= 62
= 36
Copyright © Big Ideas Learning, LLC
All rights reserved.
Chapter 6
10. (w12)5 = w12⋅5
=
1
5
1
(10y)
1
10 y
1
1000y
5
(−4)5
n
−1024
n
1024
n5
⋅
—
12. −— = —
= −—
5 =
5
1
5−1
1
5−1
10. —0 = — = —1 = −—
−1 −1 5
5
−9
⋅
( 2k1 ) = (2k1 ) = 2 (k1 ) = 32 ⋅1k ⋅ = 32k1
( 6c7 ) = (6c)7 = (6c)7 = 67c = 36c49
13. —2
14. —
5
−2
5
—
2 5
−2
—
−2
5
—
5 2 5
2
—2
—
2 5
2
—
2 2
—
10
4
62
36
−3−3
11. —
= −—3 = −— = −—
−2
3
27
3
6
—2
(−8)−2
3
()
( )
h
h 2
h2
a circle is A = πr2. Also, r = —. A = πr2 = π — = π —2 =
2
2
2
πh2
h2
π — = —.
4
4
So, two expressions that represent the area of a base of the
πh2
cylinder are πr2 and —.
4
( )
34
(−8)
81
64
12. —
= —2 = —
−4
15. The base of a cylinder is a circle. The formula for the area of
2.3 108
2.3 ×
16. —2 = — × —2
2.5 10
2.5 × 10
108
1
x
13. x−7 = —7
14. y0 = 1
9
⋅1
9
y
—
15. 9x0y−3 = —
3 = 3
y
⋅
15 1
c
15
c
16. 15c−8d 0 = —
= —8
8
2−2m−3
n
1
2m
1
4m
—3
17. —
=—
2 3=
0
= 0.92 × 106
= 0.92 × 106
= 9.2 × 105
So, Pluto orbits the Sun 9.2 × 105, or 920,000, times while
the Sun completes one orbit around the center of the
Milky Way.
⋅
⋅
4−3a0
b
⋅4 b
1
7
b7
64
p−8
7 q
72q9
p
49q9
p
—
20. —
=—
8 =
8
−2 −9
⋅1 8 y⋅ x 4 ⋅ 8y ⋅ x 32xy
⋅
13 ⋅ 1 ⋅ 5 ⋅ z
13 ⋅ 125 ⋅ z
1625z
13x y
22. — = ——
= ——
=—
x
x
x
5 z
22
1
7
7
7
21. —
=—
=—
=—
6
6
6
−1 0 −7
Vocabulary and Core Concept Check
1. Sample answer: First, use the Product of Powers Property
to simplify the expression inside the parentheses to 44.
Then, use the Power of a Power Property to simplify the
entire expression to 4−8. Then, use the definition of negative
1
1
exponents to produce the final answer, —8 = —.
65,536
4
2. Use the Power of a Product Property when a product of
factors is in parentheses, and the whole product is being
raised to a power. In order to use this property, find the
power of each factor and then multiply the powers.
3. Use the Quotient of Powers Property when powers with the
same base are being divided. In order to use the property,
find the difference of the exponents of the numerator and
denominator. The answer is the common base raised to this
difference.
⋅ This answer is
36⋅3 = 318. The other three expressions are each equal to
33+6 = 39.
36 3.”
Monitoring Progress and Modeling with Mathematics
6. 40 = 1
Copyright © Big Ideas Learning, LLC
All rights reserved.
⋅
19. —
=—
=—
3
−7
22y−6
8 zx
6.1 Exercises (pp. 296–298)
4. The one that is different is “Simplify
1
1 s
s
100r−11s —
18. —
= 2
=—
3 r11 9r11
32
= 9.2 × 10−1 × 106
5. (−7)0 = 1
1
32
1
1
(−2)
−32
1
1
2−4 —
9. —
=
=—
1 24 16
40
8. (−2)−5 = —5 = — = −—
—3
11. (10y)−3 = —3 = —
3 3 =
( 4n )
1
625
7. 5−4 = —4 = —
w60
−5 0
3
−3 −10
56
5
5
10
10
5
10
5
23. —2 = 56−2
= 54
= 625
(−6)8
(−6)
24. —5 = (−6)8−5
= (−6)3
= −216
25. (−9)2
⋅ (−9)
2
= (−9)2+2
= (−9)4
= 6561
26. 4−5
⋅4
5
= 4−5+5
= 40
=1
27. (p6)4 = p6⋅4
= p24
Algebra 1
Worked-Out Solutions
307
Chapter 6
( −t3 ) = (−t)3 = (−1)9 t = 1 ⋅9 t = t9
28. (s−5)3 = s−5⋅3
40. —
= s−15
⋅6
5
⋅
−2
( )
w3
6
= 6−3
43. −—
1
= —3
6
1
=—
216
30. −7
—
1
(3s )
1
3 (s )
1
243s
−6
( 2r1 )
44. —6
=
(−7)1
⋅
⋅
1−6
(2r 6)6 26(r 6)6
=—
=—
= — = 64
(2r 6)−6
16
1
⋅
⋅x
x = x5−4
=
⋅z
z
1
⋅ 10
5
z10
z
= 10−7+5 = 10−2
1
So, the magnified length of the object is 10−2, or —
, meter
100
(which is the same as 1 centimeter).
112a3b2
8ab
32πs6
1
(25πs6)(3−1) = (32πs6) — = —
3
3
πs
πs 32πs5 s1 32πs6
(2s)5 — = 25s5 — = — = —
3
3
3
3
32πs6
None of the other expressions are equivalent to —.
3
So, the answers are B, C, and D, because these expressions
32πs6
simplify to the volume of the sphere, which is —.
3
x2
46. t = —
2D
(10−4)2
=—
2(10−5)
1
= — 10−4⋅2 105
2
⋅
10−5 = z5
—
32. —
=—
5 = 5 =z
5
33. 10−7
112
8
34. — = — a3−1b2−1
= 14a2b microns
⋅
36. In the second step, the Quotient of Powers Property should
be used because powers with the same base are being
divided. The exponents should be subtracted, not divided.
x5
⋅x
x
3
x5+3
x
x8
x
8−4 = x4
—
=—
—
4 = 4=x
4
1
(4x)
1
4x
1
256x
—4
38. (4x)−4 = —4 = —
4 4 =
−2
( n6 )
39. —
308
6−2 n2 n2
=—
=—=—
n−2 62 36
Algebra 1
Worked-Out Solutions
5
−1
=5×
−8
5
10−1−8+5
= 5 × 10−4
It takes about 5 × 10−4, or 0.0005, second for the ink
to diffuse 1 micrometer.
(
2x−2 y3
3x y
(
4s5t−7
−2s t
47. —
−4
) (
4
37. (−5z)3 = (−5)3z3
= −125z3
⋅
⋅
= 0.5 ⋅
⋅ 10
= 5 × 10 ⋅ 10 ⋅ 10
powers with the same base are being multiplied. So, the
product should have a base of 2, not 2 2.
25 = 24+5 = 29
⋅
10−8
35. The Product of Powers Property should be used because
⋅
()
⋅
So, the length of the computer chip is 14a2b microns.
24
= 64r36
32πs6
Any expression equivalent to — represents the volume
3
of the sphere.
x
x1+1
z8+2
z
2
6 6
⋅ ⋅
⋅ ⋅
= x2
z8
⋅
⋅r ⋅
4
3
4
= — π(2s2)3
3
4
= — π 23 (s2)3
3
4
= — π 8 s6
3
32πs6
=—
3
= (−7)1+(−4)
= x1
1
243s
45. B, C, D; V = — πr3
(−7)−4
1
= —3
(−7)
1
=—
−343
1
= −—
343
x5
x
2
—
(−1)−2(w3)−2
62
36
36
= ——
=—
= —= —
6−2
(−1)2(w3)2 1 w3⋅2 w6
= (−7)−3
31. —4
2
—
42. (−5p3)3 = (−5)3(p3)3 = −125p3⋅3 = −125p9
= 6−8+5
(−7)−4
22
—
2
—
—
—
41. (3s8)−5 = —
8 5 = 5 8 5=
8⋅5 =
40
1
=—
s15
29. 6−8
2
2
48. —
−2 4
⋅ ⋅
⋅ ⋅
2 y3 y4
= —
3 x1 x2
) (
4
⋅
⋅
2 y3+4
= —
3 x1+2
) ( )
4
2y7
= —3
3x
4
(2y7)4 24(y7)4
16y7⋅4 16y28
=—
=— =—
=—
(3x3)4 34(x3)4
81x3⋅4 81x12
) (
3
⋅ ⋅
⋅ ⋅
4 s5 s2
= —
−2 t4 t7
) (
3
⋅
−2 s5+2
= —
t4+7
) ( )
3
−2s7
= —
t11
3
(−2)3(s7)3
−8s7⋅3
−8s21
(−2s7)3
= —
= —
= —
= —
11
3
11
3
33
⋅
(t )
t
t
t33
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All rights reserved.
Chapter 6
(
3m−5n2
4m n
49. —
−2 0
⋅ ⋅
⋅ m⋅
) ⋅( ) (
2
2
mn4
9n
—
) ⋅( )
( ) ⋅( )
3 m2 n2
= —
4 m53 1
3n2
= —3
4m
2
(3n2)2
=—
(4m3)2
2
mn3
9
(3.9 × 10−5)
(7.8 × 10 )
= 0.5 × 10−5−(−8)
= 0.5 × 10−5+8
= 5 × 10−1 × 103
(mn3)2
—
(32)2
= 5 × 10−1+3
32 n2⋅2 m2 n3⋅2
= ——
42 m3⋅2 32⋅2
⋅
= 5 × 102
⋅ ⋅
⋅ ⋅
3 ⋅n ⋅m ⋅n
= ——
4 ⋅m ⋅3
m 3
2
4
2
2
6
The quotient is 5 × 102, or 500.
9.46 × 10−1
2.3 × 10
6
2
≈ 4.113 × 10−1−(−5)
n4+6
≈ 4.113 × 10−1+5
= ——
42 m4 32
⋅ ⋅
≈ 4.113 × 104
n10
= —4
16 9 m
On average, about 4.113 × 104, or 41,130, pounds of
potatoes were produced for each acre harvested.
⋅ ⋅
n10
= —4
144m
( ) ⋅(
3x3y0
x
4
50. —
−2
y2 x−4
5xy
—
−8
)
3
= (3
= (3
y ⋅y
⋅ x ⋅ x ⋅ 1) ⋅ ( —
5⋅x ⋅x )
3
2
⋅
= (3x5)4
⋅( ⋅
y2+8
—
5 x1+4
y
⋅ (—
5x )
10 3
5
⋅
⋅x ⋅ ⋅—
(5x )
81 ⋅ x ⋅ y
= ——
5 x⋅
= 34
2
4
5 4
y10 3
5 3
20
30
3 5 3
x5
⋅ ⋅
x20
y30
81
= ——
125x15
81x5y30
=—
125
51. (3 × 102)(1.5 × 10−5) = 3(1.5) × (102)(10−5)
= 4.5 × 102+(−5)
= 4.5 × 10−3
The product is 4.5 × 10−3, or 0.0045.
52. (6.1 × 10−3)(8 × 109) = 6.1(8) × (10−3)(109)
= 48.8 × 10−3+9
= 4.88 × 101 × 106
= 4.88 × 101+6
= 4.88 × 107
)
3
3
8
1
x3+2)4
10−1
10
9.46
2.3
55. ——
=—×—
−5
−5
4
4
10−5
10
3.9
7.8
=—×—
54. ——
−8
−8
2
mn3
—
32
⋅
2
—
d
r
7.8 × 108
3 × 10
7.8
3
= 2.6 × 108−5
4
= 2.6 × 103
So, it takes 2.6 × 103, or 2600, seconds for sunlight to reach
Jupiter.
57. a. You should use the Power of a Product Property because
for each cube, you must raise a product of two factors to
the third power to find the volume.
(6x)3
b. Use the Power of a Quotient Property to express —3
(2x)
6x 3
as — . Simplify the expression inside the parentheses to
2x
produce (3)3, so the volume is 27 times greater.
( )
1 kilobyte
2 bytes
58. a. —
10
240 kilobytes
2 bytes
= ——
⋅—
1 terabytes 2 terabytes
40
10
= 240−10 kilobytes per terabyte
= 230 kilobytes per terabyte
So, there are 230 kilobytes in 1 terabyte.
⋅
1 megabyte 230 bytes
—
1 gigabyte
2 bytes
b. —
20
⋅
230+4
=—
megabytes
220
234
=—
megabytes
220
= 234−20 megabytes
(6.4 × 107)
(1.6 × 10 )
= 214 megabytes
6.4
1.6
= — × —5
53. —
5
10
⋅ 16 gigabytes
230 24
=—
megabytes
220
The product is 4.88 × 107, or 48,800,000.
107
108
10
— × —5
56. t = — = —
5 =
So, there are 214, or 16,384, megabytes in 16 gigabytes.
= 4 × 107−5
= 4 × 102
The quotient is 4 × 102, or 400.
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Algebra 1
Worked-Out Solutions
309
Chapter 6
c. In order to convert the number of bytes in each unit of
measure to bits, multiply each number in the table by 8.
Because 8 can be expressed as 23, multiply each number
in the table by 23. Because the values have a common
base of 2, they can be simplified using the Product of
Powers Property. So, you can simply add 3 to each of the
exponents in the table.
59.
8a3b3
=
23a3b3
=
1
= —n = —n
6
6
1
6
1
6
−2y = 2
2
−2y
—=—
−2
−2
y = −1
x − (−1) = 9
or
81x4y8 = 92x2⋅2y4⋅2 = (9x2y4)2
1
6
− 11 − 11
x−y=9
62. 81x4y8 = 34x4y2⋅4 = (3xy2)4
()
−2y + 11 = 13
Step 3:
61. 64w18z12 = 26w3⋅6 z2⋅6 = (2w3z2)6
63. a. —
x + 2 − 3y = 13
(9 + y) + 2 − 3y = 13
(2ab)3
60. 16r2s2 = 42r2s2 = (4rs)2
n
Step 2:
x+1=9
−1−1
1n
x=8
So, x = 8 and y = −1.
1
1296
b. —n = —4 = —
66. Sample answer: Let r = 9x2.
1
1
2
32
probability of flipping heads five times in a row is
1
1 5 15
— = —5 = —.
2
32
2
c. —; The probability of flipping heads once is —, and the
()
64. a.
1
V = — πr2h
3
1
27πx8 = — π(9x2)2h
3
⋅9 ⋅x ⋅ h
1
= — π ⋅ 81 ⋅ x h
3
1
27πx8 = — π
3
27πx8
2
2 2
4
27πx8 = 27πx4h
Figure 1
The shaded
part is —12 of
the original.
b.
Figure 2
The shaded
part is —14 of
the original.
Figure 4
The shaded
1
part is —
of
16
the original.
Figure 3
The shaded
part is —18 of
the original.
67. 1010(10−6) = 1010+(−6)
= 104
104 g
1
—
104 kg
1 kg
= — = 10
⋅—
10
10 g
3
3
4−3 kg
= 101 kg
So, the mass of the seed from the double coconut palm is
10 kilograms, which means your friend is incorrect.
65. Using the Quotient of Powers Property and the first
bx
equation —y = b9, you can conclude that x − y = 9.
b
Using the Product of Powers Property, the Quotient of
bx b2
= b13, you
Powers Property, and the second equation —
b3y
can conclude that x + 2 − 3y = 13. Use these equations to
solve a system of linear equations by substitution.
⋅
x−y=9
x4 = h
So, one possibility is r = 9x2 and h = x4.
1
Figure 1: — = 2−1
2
1
1
Figure 2: — = —2 = 2−2
4 2
1
1
Figure 3: — = —3 = 2−3
8 2
1
1
Figure 4: — = —4 = 2−4
16 2
Step 1:
x4
27πx4h
27πx8 = —
—4
27πx4
27πx
68. a. Each of the 13 questions has 2 choices. So, 213 represents
the number of different ways that a student can answer all
the questions in Part 1.
b. Each of the 23 questions has 2 choices. So, 223 represents
the number of different ways that a student can answer all
the questions on the entire survey.
c. Because each question now has 3 choices, the answer for
part (a) becomes 313, and the answer for part (b) becomes
323.
x−y+y=9+y
x=9+y
310
Algebra 1
Worked-Out Solutions
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Chapter 6
69. a. When a > 1 and n < 0, an < a–n because an will be less
f.
than 1 and a–n will be greater than 1. When a > 1 and
n = 0, an = a–n = 1, because any number to the zero
power is 1. When a > 1 and n > 0, an > a–n because an
will be greater than 1 and a–n will be less than 1.
—
√8
3
5
2
125
— mm = — mm, or 2.5 mm
⋅ ⋅
⋅
b. When 0 < a < 1 and n < 0, an > a–n, because an will be
greater than 1 and a–n will be less than 1. When 0 < a < 1
and n = 0, an = a–n = 1, because any number to the zero
power is 1. When 0 < a < 1 and n > 0, an < a–n because
an will be less than 1 and a–n will be greater than 1.
Of the sides measured in metric units, the longest length
is 1.5 meters. Of the sides measured in standard units, the
longest length is 1 yard. Because 1 meter is approximately
the same length as 1 yard, the cube in part (d) has the largest
side length of 1.5 meters. The cubes in parts (a) and (e) have
equal side lengths because 3 feet = 1 yard.
Maintaining Mathematical Proficiency
—
—
70. Because 52 = 25, √ 25 = √ 52 = 5.
—
71. −√ 100 represents the negative square root.
—
—
Because 102 = 100, −√100 = −√102 = −10.
4—
2. a. Sample answer: √ 25 ≈ 2.2, which is represented by point
C. 25 is between 24 = 16 and 34 = 81, and C is the only
point on the graph between 2 and 3.
—
b. Sample answer: √ 0.5 ≈ 0.7, which is represented by
point A. 0.5 is between 02 = 0 and 12 = 1, and A is the
—
only point on the graph between 0 and 1. Also, √0.5 is
—
2
√
close to 0.49 , which equals 0.7 because (0.7) = 4.9.
—
√64
1
72. ± — represents the positive or negative square root.
()
—
—
√( )
√
1
1
1 2
1
1 2
Because — = —, ± — = ± — = ±—.
64
8
8
64
8
73. 12 is a natural number, whole number, integer, rational
number, and real number.
5—
c. Sample answer: √ 2.5 ≈ 1.2, which is represented by
65
74. — is a rational number and a real number.
9
π
75. — is an irrational number and a real number.
4
6.2 Explorations (p. 299)
3—
1. a. √ 27 ft = 3 ft
point B. 2.5 is between 15 = 1 and 25 = 32, and B is the
only point on the graph between 1 and 2.
3—
d. Sample answer: √ 65 ≈ 4.0, which is represented by point
E. 65 is between 43 = 64 and 53 = 125, and E is the only
point on the graph between 4 and 5.
?
Check: 33 = 27
?
3 3 3 = 27
?
9 3 = 27
3—
b. √ 125 cm = 5 cm
3—
c. √ 3375 in. = 15 in.
3—
d. √ 3.375 m = 1.5 m
3—
e. √ 1 yd = 1 yd
Copyright © Big Ideas Learning, LLC
All rights reserved.
3—
e. Sample answer: √ 55 ≈ 3.8, which is represented by point
⋅ ⋅
⋅
27 = 27 ✓
?
Check: 53 = 125
?
5 5 5 = 125
?
25 5 = 125
⋅ ⋅
⋅
125 = 125 ✓
?
Check: 153 = 3375
?
15 15 15 = 3375
?
225 15 = 3375
⋅ ⋅
⋅
3375 = 3375 ✓
?
Check: 1.53 = 3.375
?
1.5 1.5 1.5 = 3.375
?
2.25 1.5 = 3.375
⋅ ⋅
⋅
3.375 = 3.375 ✓
?
Check: 13 = 1
?
1 1 1=1
?
1 1=1
1=1✓
⋅ ⋅
⋅
()
5 3 ? 125
Check: — = —
8
2
5 5 ? 125
5 —
—=—
—
8
2 2 2
125
25 5 ? —
— —=
8
4 2
125
125 —
✓
—=
8
8
D. 55 is between 33 = 27 and 43 = 64, and D is the only
point on the graph between 3 and 4.
6—
f. Sample answer: √ 20,000 ≈ 5.2, which is represented by
point F. 20,000 is between 56 = 15,625 and 66 = 46,656,
and F is the only point on the graph between 5 and 6.
3. Find what real number multiplied by itself n times gives
you that number. If that is not possible, determine which nth
powers the number is between and estimate the decimal part.
4.
m = (0.00016)C2.73
4000 = (0.00016)C2.73
4000
0.00016
(0.00016)C2.73
0.00016
— = ——
25,000,000 = C2.73
2.5 × 107 = C2.73
Use guess and check with a calculator:
5002.73 ≈ 2.333 × 107
5102.73 ≈ 2.464 × 107
5122.73 ≈ 2.491 × 107
512.52.73 ≈ 2.497 × 107
512.72.73 ≈ 2.500 × 107
So, the circumference of its femur was about
512.7 millimeters.
Algebra 1
Worked-Out Solutions
311
Chapter 6
6.2 Monitoring Progress (pp. 300–302)
9. The index n = 3 is odd, so 1000 has one real cube root.
1. The index n = 3 is odd, so −125 has one real cube root.
Because
= −125, the cube root of −125 is
3—
√−27 = −5, or (−125)1/3 = −5.
(−5)3
2. The index n = 6 is even, and a > 0. So, 64 has two real
sixth roots. Because
= 64 and
= 64, the sixth roots
6—
of 64 are ±√ 64 = ±2, or ±641/6 = ±2.
26
3—
(−2)6
⋅ (−5) ⋅ (−5)
3 ——
3. √ −125 = √ (−5)
= −5
Because
103 = 1000, the cube root of 1000
3—
is √ 1000 = 10, or (1000)1/3 = 10.
10. The index n = 9 is odd, so −512 has one real ninth root.
Because (−2)9 = −512, the ninth root of −512
9—
is √ −512 = −2, or (−512)1/9 = −2.
?
43 = 64
?
4 4 4 = 64
?
16 4 = 64
3—
11. √ 64 in. = 4 in.
Check:
64 = 64 ✓
4. (−64)2/3 = (−641/3)2
=(−4)2
So, each side of the cube is 4 inches.
=16
?
63 = 216
?
6 6 6 = 216
?
36 6 = 216
3—
12. √ 216 cm = 6 cm
Check:
5. 95/2 = (91/2)5
=35
=243
So, each side of the cube is 6 centimeters.
= 43
4—
( 3V4π )
1/3
(
3(17,000)
= —
4(3.14)
) (
1/3
51,000
= —
12.56
)
1/3
=4
≈ 16
3—
()
( )
1.5881/8
= −6
3—
= −7
5—
= −(4)
= −4
The annual inflation rate is about 6.0%.
7—
17. 1281/7 = √ 128
6.2 Exercises (pp. 303−304)
=2
1. Find the fourth root of 81, or what real number multiplied by
itself four times produces 81.
2— 3
2. The expression that does not belong is ( √ 27 ) because it is
the only one that is not equivalent to 9.
Monitoring Progress and Modeling with Mathematics
18. (−64)1/2 is not a real number because there is no real number
that can be multiplied by itself two times to produce −64.
—
( 5 )4 = (81/5)4
19. √ 8
= 84/5
—
√10 = 101/2
(
5— 6
20. √ −21
5—
4. √ 34 = 341/5
)
= [(−21)1/5]6
= (−21)6/5
3—
151/3 = √ 15
21. (−4)2/7 = [(−4)1/7]2
8—
6. 1401/8 = √ 140
= ( √ −4 )
7— 2
7. The index n = 2 is even and a > 0. So, 36 has two real
square roots. Because 62 = 36 and (−6)2 = 36, the square
—
roots of 36 are ±√ 36 = ±6, or ±361/2 = ±6.
8. The index n = 4 is even and a > 0. So, 81 has two real
fourth roots. Because
= 81 and
= 81, the fourth
4—
roots of 81 are ±√81 = ±3, or ±811/4 = ±3.
34
312
⋅ ⋅ ⋅ ⋅ ⋅ ⋅
7 ——
= √2 2 2 2 2 2 2
Vocabulary and Core Concept Check
5.
⋅4 ⋅4 ⋅4 ⋅4)
——
16. −√ 1024 = −( √ 4
≈ 0.0595
3.
⋅ (−7) ⋅ (−7)
3 ——
15. √ −343 = √ (−7)
−1
⋅ (−6) ⋅ (−6)
3 ——
14. √ −216 = √ (−6)
So, the radius of the beach ball is about 16 inches.
F 1/n
8. r = —
−1
P
13,500 1/8
= —
−1
8500
⋅4 ⋅4 ⋅4
4 ——
13. √ 256 = √ 4
= 64
≈
⋅ ⋅
⋅
216 = 216 ✓
6. 2563/4 = (2561/4)3
7. r = —
⋅ ⋅
⋅
Algebra 1
Worked-Out Solutions
(−3)4
22. 95/2 = (91/2)5
— 5
= ( √9 )
23. 323/5 = (321/5)3
= 23
=8
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Chapter 6
24. 1252/3 = (1251/3)2
1
6√π
1
3/2
=—
— (60)
6√π
1
≈ —(464.758)
10.6347
3/2
38. V = —
—S
= 52
= 25
25. (−36)3/2 is not a real number because
(−36)3/2 = [(−36)1/2]3, and there is no real number that can
be multiplied by itself two times to produce −36.
≈ 43.7
The volume of the sphere is about 44 cubic meters.
26. (−243)2/5 = [(−243)1/5]2
39. Write the radicand, a, as the base and write the exponent as a
= (−3)2
fraction with the power, m, as the numerator and the index, n,
as the denominator.
=9
27. (−128)5/7 = [(−128)1/7]5
40.
A = s2
= (−2)5
x = s2
= −32
—
√x
= √s2
28. 3434/3 = (3431/3)4
√x
—
=s
x1/2
=s
= 74
An expression that represents the side length of the square is
x1/2 inches.
= 2401
29. The numerator and denominator are reversed.
—
( √3 2 )4 = (21/3)4
= 24/3
30. (−81)1/4 is not a real number because there is no real number
that can be multiplied by itself four times to produce −81.
( )
1/3
1
31. —
1000
3—
√1
1
11/3
=—
=—
=—
3—
10001/3 √
1000 10
(
⋅ ⋅ ⋅ ⋅ ⋅
⋅
−1
1/n
1/10
≈ (2.41781)1/10 − 1
≈ 0.09230
The annual inflation rate is about 9.2%.
5—
5—
5—
0 = 01/5 = √ 0 ,
−1 = (−1)1/5 = √−1
So, x = x1/5 is true for x = −1, x = 0, and x = 1.
= ( √3 3 3 3 3 3 )( √ 4 )
6 ——
1/6
≈ 0.05451
43. 1 = 11/5 = √ 1 ,
= ( √729 )(41/2)
)
= (1.375)1/6 − 1
—
6—
—
n—
44. no; The value of √ a is not always positive, and the value of
—
= (3)( √ 2 2 )
—
n
−√
a is not always negative. If n is odd and a is negative,
n—
n—
then √
a will be negative and −√
a will be positive.
= (3)(2)
=6
The area of the bake sale sign is 6 square feet.
45. (y1/6)3
⋅
—
√x
⋅
= y3(1/6) x1/2
⋅
= y1/2 x1/2
3—
√275 = 275/3
=
−1
( FP ) − 1
3.53
−1
=(
1.46 )
35. A =ℓw
36.
1/n
1,100,000
= —
800,000
42. r = —
√1
11/6
1
=—
=—
=—
6—
641/6 √
2
64
1
1
1
1
33. (27)−2/3 = —
=—=—=—
272/3 ( 3 — )2 32 9
√27
1
1
1
1
34. (9)−5/2 = —
= —5 = —5 = —
95/2 ( √—
3
243
9)
1/6
( FP )
41. r = —
The annual inflation rate is about 5.5%.
6—
( 641 )
32. —
—
5
(271/3)
(
⋅
⋅
1 3 1 3
1
3 — = — — = —, or —
6 1 6 6
2
)
= (xy)1/2
—
= 35
The simplified expression is (xy)1/2, or √xy .
= 243
One side of the box is 243 millimeters.
3(5)
15
)
( 3Vπh ) = ( (3.14)(4)
) = ( 12.56
37. r = —
1/2
—
1/2
—
1/2
≈ (1.1943)1/2 ≈ 1
The radius of the paper cup is about 1 inch.
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Algebra 1
Worked-Out Solutions
313
Chapter 6
46. (y
⋅y
1/3)3/2
= (y1+1/3)3/2
true for x = 0 and x = 1.
( 1 + 31 = 33 + 13 = 43 )
= (y4/3)3/2
=
54. The statement x1/3 = x3 is sometimes true because it is only
—
y(4/3)(3/2)
(
= y2
4
3
—
—
—
—
?
01/3 = 03
—
?
3
√0 = 0 0 0
3— ?
√0 0 0 = 0 0
⋅ —23 = —6 , or 2 )
12
⋅⋅
The simplified expression is y2.
47. x
3—
⋅ √y
6
+ y2
⋅ √x
3—
3
6/3
1
2
1
2
2
3/3
1
⋅ ⋅
2
= 2xy2
The simplified expression is 2xy2.
48. (x1/3
⋅ y ) ⋅ √y = x
—
1/2 9
⋅
=
(1/2) 9
x9/3
= x3
= x3
55.
⋅y ⋅ ⋅y
⋅y ⋅y
⋅y
⋅y
(1/3) 9
9/2
1/2
1/2
(9/2)+(1/2)
10/2
= x3y5
The simplified expression is x3y5.
V = 7.66ℓ3
49.
20
7.66
⋅ ⋅
⋅
2 ≠ 512 ✗
x1/3 = x1/3
x2/3
3—
The statement —
=√
x is sometimes true. It is true by
x1/3
the Quotient of Powers Property and the definition of rational
exponents except when x = 0 because division by 0 is
undefined.
⋅x
3
x(1/3)+3
x = x(1/3)+(9/3)
2.6110 ≈ℓ3
3—
x = x10/3
3—
√2.6110 ≈ √ℓ3
1.3770 ≈ ℓ
50. Sample answer: The formula for the period of a pendulum is
—
()
ℓ 1/2
ℓ
T = 2π — , or T = 2π —
g
g
.
51. The statement (x1/3)3 = x is always true because
(x1/3)3 = x(1/3)⋅3 = x3/3 = x1 = x by the Power of a Power
Property.
x1/3
11/3
x1/3
=
x−3 is
sometimes true. If x = 1, then
1
1
=
= √1 = 1 and x−3 = 1−3 = —3 = — = 1. So, the
1
1
statement is true if x = 1. Otherwise, it is false.
52. The statement
Let x = −1.
x = x10/3
?
−1 = (−1)10/3
? 3 — 10
−1 = ( √−1 )
?
−1 = (−1)10
−1 ≠ 1 ✗
The edge length of the dodecahedron is approximately
1.38 feet.
√
1=1✓
x2/3 ? 3 —
x
?
x(2/3)−(1/3) = x1/3
x=
7.66ℓ3
7.66
—=—
⋅⋅
= √x
—
1/3
56. x = x1/3
20 = 7.66ℓ3
0=0✓
⋅ ⋅
⋅
?
81/3 = 83
3— ?
√8 = 8 8 8
3— ?
√2 2 2 = 64 8
⋅y +y ⋅x
=x ⋅y +y ⋅x
= 2(x ⋅ y )
= x1
⋅ ⋅
⋅
?
11/3 = 13
—
?
3
√1 = 1 1 1
3— ?
√1 1 1 = 1 1
3—
3—
53. The statement x1/3 = √ x is always true because of the
Let x = 0.
Let x = 1.
x=
?
0 = 010/3
? 3 — 10
0 = ( √0 )
?
0 = 010
x = x10/3
?
1 = 110/3
? 3 — 10
1 = ( √1 )
?
1 = 110
0=0✓
1=1
x10/3
Let x = 8.
x = x10/3
?
8 = 810/3
? 3 — 10
8 = ( √8 )
?
8 = 210
⋅
So, the statement x = x1/3 x3 is
sometimes true. If x = 0 or x = 1,
the statement is true. Otherwise
it is false.
8 ≠ 1024 ✗
definition of a rational exponent.
314
Algebra 1
Worked-Out Solutions
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Chapter 6
Maintaining Mathematical Proficiency
f (x) = 2x − 10
f (x) = 2x − 10
1.
f (0) = 2(0) − 10
⤻ ⤻ ⤻ ⤻⤻
f (−3) = 2(−3) − 10
= −6 − 10
= 0 − 10
+1
= −16
= −10
+1
f (x) = 2x − 10
+1
f (8) = 2(8) − 10
+1
= 16 − 10
+1
=6
x
16(2)x
y
0
16(2)0
16
1
16(2)1
32
2
16(2)2
64
3
16(2)3
128
4
16(2)4
256
5
16(2)5
512
x
16(2)x
y
0
16(2)0
16
2
16(2)2
64
4
16(2)4
256
6
16(2)6
1024
8
16(2)8
4096
10
16(2)10
16,384
⤻⤻ ⤻ ⤻ ⤻
57.
6.3 Explorations (p. 305)
×2
×2
×2
×2
×2
58.
w(x) = −5x − 1
w(x) = −5x − 1
w(0) = −5(0) − 1
= 15 − 1
=0−1
= 14
= −1
⤻ ⤻ ⤻ ⤻⤻
w(−3) = −5(−3) − 1
+2
+2
+2
w(x) = −5x − 1
+2
w(8) = −5(8) − 1
+2
= −40 − 1
= −41
h(x) = 13 − x
h(−3) = 13 − (−3)
= 13 + 3
×4
×4
×4
×4
×4
Each value of x increases by the same amount, while each
value of y is multiplied by the same factor.
So, w(−3) = 14, w(0) = −1, and w(8) = −41.
59.
⤻⤻ ⤻ ⤻ ⤻
So, f (−3) = −16, f (0) = −10, and f (8) = 6.
h(x) = 13 − x
h(x) = 13 − x
h(0) = 13 − 0
h(8) = 13 − 8
= 13
=5
= 16
So, h(−3) = 16, h(0) = 13, and h(8) = 5.
60.
g(x) = 8x + 16
g(x) = 8x + 16
g(x) = 8x + 16
g(−3) = 8(−3) + 16 g(0) = 8(0) + 16 g(8) = 8(8) + 16
= −24 + 16
= 0 + 16
= 64 + 16
= −8
= 16
= 80
So, g(−3) = −8, g(0) = 16, and g(8) = 80.
Copyright © Big Ideas Learning, LLC
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Algebra 1
Worked-Out Solutions
315
Chapter 6
1
16 —
2
0
—
⤻ ⤻ ⤻ ⤻ ⤻
0
+2
2
+2
4
+2
6
+2
8
+2
10
2
4
3
2
4
1
5
—
5
x
8
—
4
+1
1
—
3
+1
16
—
2
+1
0
—
1
+1
1
2
—
—
—
—
—
—
1
×—
2
16,000
0
2
4
6
2x
2−2
20
22
24
26
y
—
1
4
1
4
16
64
1
×—
2
y
60
40
1
×—
2
1
×—
2
y = 2x
20
−2
2
y
2(3)x
0
2
4
6
2(3)−2
2(3)0
2(3)2
2(3)4
2(3)6
2
9
2
18
162
1458
0
16
2
4
4
1
6
1
4
—
8
1
16
—
10
1
—
64
1
×—
4
1600
12
8
8000
y = 16(2)x
y = 2(3)x
800
1
×—
4
1
×—
4
12,000
y
1200
1
×—
4
16
—
y
1
×—
4
6 x
4
−2
b. x
x
y
4000
−2
80
400
−2
The statement seems to be true because as the exponent
increases by a constant amount, the base is multiplied by
itself the same number of additional times.
3.
x
1
×—
2
—
()
1
16( )
2
1
16( )
2
1
16( )
2
1
16( )
2
1
16( )
2
1
16( )
2
1
16 —
2
5. a.
y
⤻ ⤻ ⤻ ⤻ ⤻
⤻ ⤻ ⤻ ⤻ ⤻
+1
x
()
1
16( )
2
1
16( )
2
1
16( )
2
1
16( )
2
1
16( )
2
1
16( )
2
x
⤻ ⤻ ⤻ ⤻ ⤻
2.
2
4
−2
c. x
6 x
0
2
4
6
3(1.5)x 3(1.5)−2 3(1.5)0 3(1.5)2 3(1.5)4 3(1.5)6
y
1.3
3
6.75
15.188 34.172
y
40
y
y = 3(1.5)x
1 x
y = 16( 2)
30
20
4
10
−4
4
8
12
16 x
−4
4
8
12 x
−2
2
4
6 x
Both are curved and do not intersect the x-axis; The
graph from Exploration 1 is increasing, the graph from
Exploration 2 is decreasing.
4. Sample answer: All graphs of an exponential function
seem to have a similar curved shape, and they do not
intersect the x-axis.
316
Algebra 1
Worked-Out Solutions
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Chapter 6
d.
−6
−4
−6
−4
x
x
( 12 ) ( 21 )
—
—
= 26
( 21 )
—
64
16
x
−2
0
x
−2
() ()
1
—
2
y
=
22
4
1 x
y = ( 2)
−6
3
2 —
4
y
0
1
—
2
2
−2
() ()
x
3
2 —
4
y
1
4
—
y = 2(
)
e.
−6
2 x
−6
−6
( ) 3( 21 )
1
3 —
2
—
y
()
1
= 3(64) 3 —
2
x
−2
( ) 3( 12 )
—
y
1 x
y = 3( 2)
0
—
3
0
1
2
3
y
8
4
2
1
⤻
⤻
⤻
1
1
1
2
×—
2
—
150
−2
×—
2
×—
2
As x increases by 1, y is multiplied by —12 . So, the function is
exponential.
0.75
⤻
+4 ⤻
+4 ⤻
+4
2.
y
2 x
x
x
−4
0
4
8
y
1
0
−1
−2
⤻ ⤻ ⤻
+(−1) +(−1) +(−1)
50
−4
1.5
y
−2
100
−6
2
2
—2
⤻
+1 ⤻
+1 ⤻
+1
2
( ) = 3(1) 3( 12 ) = 3( 41 )
12
−4
1.
0
1
= 3(4) 3 —
2
200
( ) = 2( 43 )
6.3 Monitoring Progress (pp. 306–309)
= 3(16)
48
−2
1
3 —
2
= 2(1)
2
These graphs have the same characteristics as the graphs
from Exploration 3. They have the same general curved
shape, and they do not intersect the x-axis.
−4
−4
192
x
2
3
2 —
4
4
−2
x
0
8
60
x
( ) ()
3
2 —
4
12
20
−4
0
42
= 2 —2
3
16
y
( )
44
= 2 —4
3
6.3
3.6
3 x
4
40
−6
( ) ()
−4
3
2 —
4
−2
3
2 —
4
1
= —2
2
−4
46
= 2 —6
3
11.2
x
2
1
80
x
() ()
( ) =1 ( )
1
—
2
−6
x
3
2 —
4
= 24
y
1
—
2
f.
2 x
As x increases by 4, y decreases by 1. The rate of change is
constant. So, the function is linear.
3. y = 2(9)x
= 2(9)−2
( )
1
=2 —
81
2
=—
81
4. y = 1.5(2)x
y = 2(9)x
y = 2(9)x
= 2(9)0
= 2(9)12
= 2(1)
= 2( √ 9 )
=2
= 2(3)
y = 1.5(2)x
—
y = 1.5(2)x
= 1.5(2)−2
= 1.5(2)0
= 1.5(2)12
= 1.5(0.25)
= 1.5(1)
= 1.5√2
= 0.375
= 1.5
≈ 1.5(1.4142)
—
≈ 2.12
Copyright © Big Ideas Learning, LLC
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Algebra 1
Worked-Out Solutions
317
Chapter 6
−2
5. x
−1
0
−2(4)x −2(4)−2 −2(4)−1 −2(4)0
1
1
−—
−2
f (x)
−—
8
2
1
2
−2(4)1
−2(4)2
−8
−32
7. x
−2(3)−2
−1
−2(3)−1 −1
5
−—
3
−2
−1
0
−2(3)0 −1
−2(3)1 −1
−2(3)2 −1
−3
−7
−19
11
−—
9
x
−2
−2(3)x+2 − 1
4 x
2
y
f(x) = −2(4)x
−8
−16
y
−24
−6
−32
−2
x
−1
−2
() ()
1
2 —
4
1
2 —
4
f (x)
32
0
−1
()
1
2 —
4
32
f(x) = 2(
−4
−2
)
2 x
−20
8
1
()
1
2 —
4
2
0
1
2
—
−40
From the graph, you can see that the domain is all real
numbers and the range is y < −1.
2
()
1
2 —
4
−30
1
()
1
2 —
4
2
8.
−2
x
1
8
—
y
f (x)
x
0
(0.25)x
(0.25)−2
+3
+3
1
(0.25)0
f (x)
+3
40
8
30
2
+3
3.25
(0.25)1
4
16
2
−1
+ 3 (0.25)−1 + 3
19
7
(0.25)x
24
1 x
4
−4
y = −2(3)x+2 − 1
The parent function is g(x) = 4x. The graph of f is a vertical
stretch by a factor of 2 and a reflection in the x-axis of the
graph of g. The y-intercept of the graph of f, −2, is below the
y-intercept of the graph of g, 1. From the graph of f, you can
see that the domain is all real numbers and the range is y < 0.
6. x
−3
−1
y
y
−4
−4
−2(3)x+2
(0.25)2 + 3
3.0625
y
20
4 x
y = (0.25)x + 3
10
x
()
1
The parent function is g(x) = — . The graph of f is a
4
vertical stretch by a factor of 2 of the graph of g. The
y-intercept of the graph of f, 2, is above the y-intercept of
the graph of g, 1. From the graph of f, you can see that the
domain is all real numbers and the range is y > 0.
−4
−2
2
4 x
From the graph, you can see that the domain is all real
numbers and the range is y > 3.
9.
x
−2
−1
0
1
2
g(x)
0.4
1.3
4
12
36
2
4 x
40
y
30
20
10
−4
−2
Both functions have the same value when x = 0, but the value
of f is less than the value of g over the rest of the interval.
318
Algebra 1
Worked-Out Solutions
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Chapter 6
10. a. Because the graph crosses the y-axis at (0, 100), the
y-intercept is 100. Also, the y-values increase by a factor
200
of — = 2 as x increases by 1.
100
because it fits the pattern y = abx, where a = −1 and b = 3.
9. The equation y = 9(−5)x does not represent an exponential
function. Although it fits the pattern y = abx, the definition of
an exponential function states that b cannot be negative.
abx
y = 100(2)x
1
2
function. Although it fits the pattern y = abx, the definition of
an exponential function states that b cannot be 1.
10. The equation y = —(1)x does not represent an exponential
b. y = 100(2)x
= 100(2)6
= 100(64)
11.
= 6400
⤻⤻⤻
+1
+1
+1
So, there are 6400 bacteria after 6 days.
c. The bacteria population in Example 7 is growing by a
factor of 4, and the bacteria population in this problem is
only growing by a factor of 2. So, this bacteria population
does not grow faster.
12.
1. Sample answer:
+1
+1
+1
⤻⤻⤻
8
y
1
2
3
4
−2
0
2
4
+2
+2
+2
As x increases by 1, y increases by 2. The rate of change is
constant. So, the function is linear.
6.3 Exercises (pp. 310–312)
Vocabulary and Core Concept Check
x
⤻⤻⤻
So, the population can be modeled by y = 100(2)x.
y
6
x
1
2
3
4
y
6
12
24
48
×2
×2
×2
⤻⤻⤻
y=
8. The equation y = −3x represents an exponential function
As x increases by 1, y is multiplied by 2. So, the function
is exponential.
4
2
+1
⤻
13.
−4
−2
2
4 x
x
y
2. The y-intercept occurs when x = 0. So, when x = 0, the
−1
0.25
⋅
⤻
x
y
6. The equation y = −6x does not represent an exponential
function because it fits the pattern y = mx + b, and therefore
represents a linear function.
7. The equation y = 2x3 does not represent an exponential
function because the exponent is a constant.
Copyright © Big Ideas Learning, LLC
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⤻
2
16
×4
⤻
3
64
×4
−3
10
⤻
+3
⤻
0
1
⤻
+3
⤻
3
−8
⤻
+3
⤻
6
−17
⤻
9
−26
As x increases by 3, y decreases by 9. The rate of change,
−9
—, or −3, is constant. So, the function is linear.
3
15. y = 3x
y = 32 = 9
5. The equation y = 4(7)x represents an exponential function
where a = 4 and b = 7.
1
4
+1
⤻
+(−9) +(−9) +(−9) +(−9)
the base of the power is a negative number, this is not an
exponential function. The other three equations represent
exponential functions.
because it fits the pattern y =
×4
+3
⤻
14.
4. The equation that does not belong is f (x) = (−3)x. Because
abx,
⤻
+1
⤻
As x increases by 1, y is multiplied by 4. So, the function is
exponential.
3. The graph of y = 5x is the parent function for the graph
Monitoring Progress and Modeling with Mathematics
0
1
×4
value of the function is y = ab0 = a 1 = a.
of y = 2(5)x. The graph of y = 2(5)x is a vertical stretch
by a factor of 2 of the graph of y = 5x. The y-intercept of
y = 2(5)x, 2, is above the y-intercept of y = 5x, 1. They both
have a domain of all real numbers and a range of y > 0.
+1
⤻
16.
f (x) = 3(2)x
f (−1) = 3(2)−1
()
1
=3 —
2
3
=—
2
Algebra 1
Worked-Out Solutions
319
Chapter 6
17. y = −4(5)x
f (x)
= −100
12
f (x) = 0.5x
16
f (−3) = 0.5−3
()
1
= —
2
=
−1
3(0.5)x 3(0.5)−2 3(0.5)−1
= −4(25)
18.
−2
25. x
y = −4(5)2
6
0
1
2
3(0.5)0
3(0.5)1
3
—
2
3(0.5)2
3
—
4
3
y
12
−3
8
23
f(x) = 3(0.5)x
=8
−4
1
3
1
1
f (3) = —(6)3 = —(216) = 72
3
3
19. f (x) = —(6)x
2
4 x
The parent function is g(x) = (0.5)x. The graph of f is
a vertical stretch by a factor of 3 of the graph of g. The
y-intercept of the graph of f, 3, is above the y-intercept of
the graph of g, 1. From the graph of f, you can see that the
domain is all real numbers and the range is y > 0.
1
4
1
y = —(4)32
4
1
= —(412)3
4
1 — 3
= —( √4 )
4
1 3
= —(2)
4
1
= —(8)
4
20. y = —(4)x
−2
−1
0
1
2
−4−2
1
−—
16
−4−1
1
−—
4
−40
−41
−42
−1
−4
−16
26. x
−4x
f (x)
y
−4
−2
2
−4
=2
21. C; The parent function of f (x) =
−2
4 x
f(x) = −4x
−8
is g(x) =
The graph of the parent function, g, decreases as x increases
because 0 < b < 1. The graph of f is a vertical stretch of the
graph of g, and the y-intercept of f is 2 because a = 2. So,
the function f matches graph C.
2(0.5)x
(0.5)x.
22. B; The parent function of f (x) = −2(0.5)x is g(x) = (0.5)x.
The graph of the parent function, g, decreases as x increases
because 0 < b < 1. The graph of f is a vertical stretch and a
reflection in the x-axis of the graph of g, and the y-intercept of
f is −2 because a = −2. So, the function f matches graph B.
−12
−16
The parent function is g(x) = 4x. The graph of f is a reflection
in the x-axis of the graph of g. The y-intercept of the graph
of f, −1, is below the y-intercept of the graph of g, 1. From
the graph of f, you can see that the domain is all real numbers
and the range is y < 0.
23. A; The parent function of f (x) = 2(2)x is g(x) = (2)x. The
graph of the parent function, g, increases as x increases
because b > 1. The graph of f is a vertical stretch of the
graph of g, and the y-intercept of f is 2 because a = 2. So,
the function f matches graph A.
24. D; The parent function of f (x) = −2(2)x is g(x) = (2)x. The
graph of the parent function, g, increases as x increases
because b > 1. The graph of f is a vertical stretch and a
reflection in the x-axis of the graph of g, and the y-intercept
of f is −2 because a = −2. So, the function f matches
graph D.
320
Algebra 1
Worked-Out Solutions
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Chapter 6
−2
27. x
−1
0
−2(7)x −2(7)−2 −2(7)−1 −2(7)0
2
−—
7
2
−—
49
f (x)
−2
1
2
−2(7)1
−2(7)2
−14
−98
−2
−1
0
1
2
—(8)x
—(8)−2
—(8)−1
—(8)0
—(8)1
—(8)2
f (x)
—
1
128
—
1
16
—
1
2
4
32
29. x
1
2
1
2
y
−4
−2
32
4 x
2
−80
16
−120
8
−160
−4
The parent function is g(x) = 7x. The graph of f is a vertical
stretch by a factor of 2 and a reflection in the x-axis of the
graph of g. The y-intercept of the graph of f, −2, is below
the y-intercept of the graph of g, 1. From the graph of f, you
can see that the domain is all real numbers and the range is
y < 0.
28. x
−2
()
1
6 —
3
x
−1
−2
()
1
6 —
3
f (x)
54
80
−1
0
()
1
6 —
3
18
6
1
6 —
3
1
2
1
2
() () ()
0
1
6 —
3
2
1
1
6 —
3
2
—
3
2
f(x) = 2(8)x
1
4 x
2
The parent function is g(x) = 8x. The graph of f is a vertical
1
shrink of the graph of g by a factor of —. The y-intercept of
2
1
the graph of f, —, is below the y-intercept of the graph of g, 1.
2
From the graph of f, you can see that the domain is all real
numbers and the range is y > 0.
30. x
−2
−1
0
1
2
3
3
3
3
3
3
x
−2
−1
0
1
—(0.25) —(0.25)
—(0.25)
—(0.25) —(0.25) —(0.25)2
2
2
2
2
2
2
3
3
3
f (x)
24
6
—
—
—
2
8
32
y
32
y
24
40
16
f(x) = 6(
−2
1
2
y
−2
60
−4
1
2
24
f(x) = −2(7)x
−40
1
2
2
1 x
3
)
8
4 x
−4
()
−2
f(x) = 2 (0.25)x
3
2
4 x
x
1
The parent function is g(x) = — . The graph of f is a
3
vertical stretch by a factor of 6 of the graph of g. The
y-intercept of the graph of f, 6, is above the y-intercept of
the graph of g, 1. From the graph of f, you can see that the
domain is all real numbers and the range is y > 0.
Copyright © Big Ideas Learning, LLC
All rights reserved.
The parent function is g(x) = (0.25)x. The graph of f is a vertical
3
stretch of the graph of g by a factor of —. The y-intercept of the
2
3
graph of f, —, or 1.5, is above the y-intercept of the graph of
2
g, 1. From the graph of f, you can see that the domain is all
real numbers and the range is y > 0.
Algebra 1
Worked-Out Solutions
321
Chapter 6
31.
−2
x
3x
−1
3−2
−1
−1
8
−—
9
f (x)
3−1
−1
2
−—
3
0
1
−1
30
31
0
2
−1
2
32
34.
−1
−3
x
()
x+1
1
− —
2
8
y
()
−2
1
−3 − —
2
−2
()
1
−3 − —
2
−1
y
−7
−5
x
−1
0
−3
12
8
()
4
1
− —
2
f(x) = 3x − 1
4 x
2
x+1
()
()
1 0
−3 − — −3
2
()
1 1
− — −3
2
1 2
− — −3
2
−3.5
−3.25
−4
y
From the graph, you can see that the domain is all real
numbers and the range is y > −1.
1
y
32.
x
−5
−4
−3
−2
−1
4x+3
4−2
1
—
16
4−1
1
—
4
40
41
42
1
4
16
f (x)
−6
−4
1 x+1
y = −( 2)
−2
2 x
−3
−8
−12
y
−16
f(x) = 4 x+3
From the graph, you can see that the domain is all real
numbers and the range is y < −3.
16
8
35.
−6
−4
−2
2 x
From the graph, you can see that the domain is all real
numbers and the range is y > 0.
33. x
0
1
2
3
4
5x−2 + 7 5−2 + 7 5−1 + 7 50 + 7 51 + 7 52 + 7
y
7.04
32
−4
x
7.2
8
12
32
−8(0.75)x+2
−8(0.75)−2
−2
y
−16.22
x
−2
−8(0.75)x+2 − 2
−3
− 2 −8(0.75)−1 − 2
−12.67
−1
0
−8(0.75)0 − 2 −8(0.75)1 − 2 −8(0.75)2 − 2
y
−10
y
−8
−6.5
y
24
−4
16
−2
2
−4
y = 5x−2 + 7
−2
2
4
6 x
From the graph, you can see that the domain is all real
numbers and the range is y > 7.
322
4 x
Algebra 1
Worked-Out Solutions
y = −8(0.75)x+2 − 2
−12
−16
From the graph, you can see that the domain is all real
numbers and the range is y < −2.
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All rights reserved.
Chapter 6
36.
−1
0
3(6)−2 − 5
3(6)−1 − 5
x
3(6)x−1 − 5
f (x)
−4.917
−4.5
−1
1
2
3.2
8
20
50
f (x)
0.125
0.5
2
8
2
3
80
3(6)0 − 5
3(6)1 − 5
3(6)2 − 5
60
−2
13
103
40
f (x)
0
g(x)
1
x
3(6)x−1 − 5
43. x
y
20
18
y
−4
−2
4 x
2
12
The value of f is less than the value of g over the entire
interval.
6
−4
−2
2
−6
4 x
f(x) = 3(6)x−1 − 5
From the graph, you can see that the domain is all real
numbers and the range is y > −5.
44. x
0
1
2
3
4
5
h(x)
32
16
8
4
2
1
f (x)
0.5
2
8
32
128
512
32
1
37. The graph of g is a vertical shrink by a factor of — of the
2
1
graph of f. So, a = —.
2
y
24
16
38. The graph of g is a vertical translation 3 units up of the graph
of f. So, k = 3.
39. The graph of g is a horizontal translation 4 units right of the
8
−2
2
6 x
4
graph of f. So, h = 4.
40. The graph of g is a horizontal translation 2 units left of the
graph of f. So, h = −2.
Both functions have the same value when x = 2, but the
value of h is greater than the value of f over the rest of the
interval.
41. According to the order of operations, the power should be
simplified before multiplying by 6.
g(x) = 6(0.5)x; x = −2
g(−2) = 6(0.5)−2
45. a. x
0.25x
y
= 6(4)
Portion of screen display
The domain is all real numbers and the range is y < −1.
1
2
3
0.250
0.251
0.252
0.253
1
0.25
0.0625
0.015625
Zoom Display
= 24
42. The graph approaches the line y = −1, not y = 0.
0
y
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
y = 0.25x
0 1 2 3 4 5 x
Zooms
Based on the context of the problem, because you cannot
zoom a negative number of times, the domain is x ≥ 0.
From the graph, you can see that the range is
0 < y ≤ 1.
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Algebra 1
Worked-Out Solutions
323
Chapter 6
b. The y-intercept is 1. This means that when you do not
zoom in, 100%, or all, of the original screen display is
seen.
+1 ⤻
+1 ⤻
+1
⤻
48.
c. y = 0.25x
x
0
1
y
−50
−10
y = 0.252
= 0.0625
46. a. x
15(3)x
0
1
2
3
4
15(3)0
15(3)1
15(3)2
15(3)3
15(3)4
15
45
135
405
1215
y
Population
Coyote Population
y
1200
1050
900
750
600
450
300
150
0
−0.4
×—
5
×—
5
When x = 0, the value of y is −50. So, the y-intercept
1
is −50. Each y-value is multiplied by a factor of — as x
5
1
increases by 1. Using a = −50 and b = —, an exponential
5
function of the form y = abx that is represented by the table
x
1
is y = −50 — .
5
()
+1 ⤻
+1 ⤻
+1
⤻
49.
x
0
1
y
−0.5
−1
2
3
−2
−4
⤻ ⤻ ⤻
×2
×2
×2
The graph crosses the y-axis at (0, −0.5). So, the y-intercept
is −0.5. The y-values increase by a factor of 2 as x increases
by 1. Using a = −0.5 and b = 2, an exponential function
of the form y = abx that is represented by the table is
y = −0.5(2)x.
y = 15(3)x
0 1 2 3 4 x
Twenty-year
periods
Based on the context of the problem, because you cannot
have a negative number of 20-year periods, the domain is
x ≥ 0. From the graph, you can see that the range is
y ≥ 15.
b. The y-intercept is 15. This means that the coyote
population was 15 at the beginning of the first interval.
c. y = 15(3)x
+1
+1 ⤻
+1
⤻
⤻
50.
x
0
1
2
3
y
8
4
2
1
⤻
⤻
⤻
1
1
1
×—
2
y = 15(3)2
= 15(9)
= 135
In 40 years, 2 twenty-year periods have passed. So, there
will be 135 coyotes in 40 years.
+1
+1 ⤻
+1
⤻
⤻
47.
−2
3
⤻
⤻
⤻
1
1
1
×—
5
So, you see 6.25% of the original screen if you zoom
in twice.
2
x
0
1
2
3
y
2
14
98
686
×—
2
×—
2
The graph crosses the y-axis at (0, 8). So, the y-intercept
1
is 8. Each y-value is multiplied by a factor of — as x
2
1
increases by 1. Using a = 8 and b = —, an exponential
2
function of the form y = abx that is represented by the
x
1
graph is y = 8 — .
2
()
⤻ ⤻ ⤻
×7
×7
×7
When x = 0, the value of y is 2. So, the y-intercept is 2. The
y-values increase by a factor of 7 as x increases by 1. Using
a = 2 and b = 7, an exponential function of the form
y = abx that is represented by the table is y = 2(7)x.
324
Algebra 1
Worked-Out Solutions
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Chapter 6
+1
+1 ⤻
+1
⤻
⤻
51. a.
x
0
1
2
3
y
40
60
90
135
54. Your friend is incorrect. This is not an exponential function
because even though the values of y are being multiplied
by a common factor, the values of x are not increasing at a
constant rate.
⤻
⤻
⤻
3
3
3
×—
2
×—
2
55. When a is positive, it causes a vertical stretch (for a > 1) or
×—
2
When x = 0, the value of y is 40. So, the y-intercept is 40.
3
The y-values increase by a factor of — as x increases by 1.
2
3
Using a = 40 and b = —, an exponential function of the
2
x that is represented by the graph is
form y = ab
x
3
y = 40 — .
2
x
3
b. y = 40 —
2
3 5
y = 40 —
2
= 40(7.59375)
()
()
()
= 303.75
So, after 5 months, the number of visitors is about 304.
52. The y-intercept is 3300 because that is the initial value.
The y-values increase by a factor of 1 + 6% = 1.06 as
the number of years x increases by 1. Using a = 3300 and
b = 1.06, an exponential function of the form y = abx that
represents this situation is y = 3300(1.06)x.
y = 3300(1.06)x
shrink (for 0 > a > 1) of the graph of the parent function. If
a is negative, it causes a vertical stretch (for a > 1) or shrink
(for 0 > a > 1) and a reflection in the x-axis of the graph of
the parent function.
56. Sample answer: The equation f(x) = 4x−5 represents a
horizontal translation 5 units right of the graph of h(x) = 4x.
57. Using the form y = abx−h + k, where h = 3 and k = 4, an
equation is g(x) = 5x−3 + 4.
58. a. The point on the graph with a y-value of 20 has an x-value
of 2. So, the stock will be worth $20 after 2 weeks.
b. The stock price in Week 1 is $40, and the stock price in
Week 3 is $10. So, the stock price drops $40 − $10 = $30
from Week 1 to Week 3.
59. The graph crosses the y-axis at (0, −1.5). So, the y-intercept
is −1.5. Each y-value is multiplied by a factor of
−6
−3
— = — = 2 as x increases by 1. Using a = −1.5 and
−3
−1.5
b = 2, an exponential function of the form y = abx that is
represented by the graph is f(x) = −1.5(2)x.
f (x) = −1.5(2)x
y = 3300(1.06)6
f (7) = −1.5(2)7
≈ 3300(1.4185)
= −1.5(128)
≈ 4681
= −192
The store expects to sell about 4681 grills in Year 6.
So, f (7) = −192.
53.
x
f (x) =
−2x
g(x) = −2x − 3
−2
−1
0
1
2
1
−—
4
1
−—
2
−1
−2
−4
1
−3—
4
1
−3—
2
−4
−5
−7
y
−4
−2
2
4 x
f(x) = −2 x
g(x) = −2 x − 3
−8
60. Sample answer: If you were to do something nice for
5 people and then ask each of those people to do something
nice for 2 more people who will each do something nice for
2 more people and so on, this situation can be modeled by
the exponential function y = 5(2)x. The y-intercept is a = 5
because initially 5 people had someone do something nice
for them, and b = 2 because the number of people who
have someone do something nice for them is increasing by a
common factor of 2.
f(x + k)
f(x)
abx+k
ab
−12
−16
bx+k
b
(x+k)−x = bk
—
61. — = —
x =
x =b
4
2
that represents this situation is f(x) = 5(2)x.
62. Using a = 5 and b = — = 2, an exponential function of the
form y =
abx
The y-intercept of g is 3 units below the y-intercept of f. The
domain of both functions is all real numbers. The range of g
is y < −3 and the range of f is y < 0.
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Algebra 1
Worked-Out Solutions
325
Chapter 6
63. Sample answer:
y = 1188(1.5)t
Let f(0) = 8. So, a = 8.
100,000 = 1188(1.5)t
Use the equation for slope. Solve for f(2).
2
⋅
100,000
1188
84.175 ≈ (1.5)t
Use a table of values, or Guess, Check, and Revise to find
that if t = 11, then 1.511 ≈ 86.498, which is close to 84.175.
So, the population will return to about 100,000 nesting pairs
after 11 intervals of 5-years, or about 55 years after 1981,
which is the year 2036.
⋅
24 = f(2) − 8
+8
+8
32 = f(2)
2. a. 80.5 − 60 = 20.5
78.5 − 60 = 18.5
Use the exponential form of a function. Solve for b.
f(x) = a(b)x
2
20.5 − 18.5
— = — ≈ 0.0976, or about 9.8%
20.5
20.5
32 = 8(b)2
32 = 8b2
8
b. Time
8
4 = b2
—
√4 =
√b2
2=b
Using a = 8 and b = 2, an exponential function for this
situation is f(x) = 8(2)x.
Maintaining Mathematical Proficiency
4
100
64. 4% = — = 0.04
35
100
128
66. 128% = — = 1.28
100
250
67. 250% = — = 2.5
100
65. 35% = — = 0.35
Temperature
difference (°F)
Body temperature
(°F)
0
98.6 − 60 = 38.6
98.6
1
38.6(1 − 0.098) ≈ 34.8
60 + 34.8 = 94.8
2
34.8(1 − 0.098) ≈ 31.4
60 + 31.4 = 91.4
3
31.4(1 − 0.098) ≈ 28.3
60 + 28.3 = 88.3
4
28.3(1 − 0.098) ≈ 25.5
60 + 25.5 = 85.5
5
25.5(1 − 0.098) ≈ 23.0
60 + 23.0 = 83.0
6
23.0(1 − 0.098) ≈ 20.7
60 + 20.7 = 80.7
So, the time of death was about 6 hours prior to midnight,
or about 6 p.m.
3. As the independent variable changes by a constant amount,
the dependent variable is multiplied by a constant factor.
4. a. Sample answer: the value of a CD each year that earns
2.5% interest compounded annually
6.4 Explorations (p. 313)
+5
+5 ⤻
+5 ⤻
+5 ⤻
+5
⤻
⤻
x
1981
1986
1991
1996
2001
2006
y
1188
1875
3399
5094
6846
9789
⤻ ⤻ ⤻⤻⤻
×1.58
×1.81 ×1.50 ×1.34 ×1.43
As x increases by 5, y is multiplied by about 1.5. So, this
situation can be described as exponential growth. Using
a = 1188 and b = 1.5, an exponential function of the form
y = abt that can approximately model this situation is
y = 1188(1.5)t, where t represents the number of 5-year
intervals since 1981.
326
(h)
—
1.
1188(1.5)t
1188
— =—
f(2) − f(0)
m=—
2−0
f(2) − 8
12 = —
2
f(2) − 8
12 = 2 —
2
Algebra 1
Worked-Out Solutions
b. Sample answer: the worth of a farm tractor that depreciates
at a rate of 10% per year
6.4 Monitoring Progress (pp. 314–318)
1. a. The initial amount is 500,000 and the rate of growth is
15%, or 0.15.
y = a(1 + r)t
y = 500,000(1 + 0.15)t
y = 500,000(1.15)t
The website membership can be represented by
y = 500,000(1.15)t.
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Chapter 6
b. The value t = 6 represents 2016 because t = 0 represents
2010.
= (0.95)t
y = 500,000(1.15)t
≈
y = 500,000(1.15)6
≈ 1,156,530
So, in 2016, the website will have about
1,160,000 members.
⤻
+1 ⤻
+1 ⤻
+1
0
1
2
3
y
64
16
4
1
)
(
(
⤻
⤻
⤻
1
1
1
)
x
1
3
5
7
y
4
11
18
25
⤻
⤻⤻
+7 +7 +7
As x increases by 2, y increases by 7. The function has a
constant rate of change. So, it is a linear function and
therefore neither an exponential growth nor an exponential
decay function.
4. The function is of the form y = a(1 −
where 1 − r < 1.
So, it represents exponential decay. Use the decay factor
1 − r to find the rate of decay.
r)t,
1 − r = 0.92
−1
−r = −0.08
−0.08
−r —
—=
−1
−1
r = 0.08
So, the rate of decay is 8%.
5. The function is of the form y = a(1 + r)t, where 1 + r > 1.
So, it represents exponential growth. Use the growth factor
1 + r to find the rate of growth.
1 + r = 1.2
−1
r = 0.2
0
500
2
4
6
8
598.21 715.70 856.28 1024.5
Savings Account
Balance (dollars)
⤻
+2 ⤻
+2 ⤻
+2
−1
2
So, the function represents exponential decay.
r nt
8. y = P 1 + —
n
0.09 12t
y = 500 1 + —
12
t
y
×— ×— ×—
4
4
4
As x increases by 1, y is multiplied by —14. So, the table
represents an exponential decay function.
−1
⋅ (0.95)
⋅ 0.90
y = 500(1.0075)12t
x
3.
(0.95)t
≈ 0.9(0.95)t
≈ 500,000(2.313)
2.
7. y = (0.95)t+2
y
1200
1050
900
750
600
450
300
150
0
y = 500(1.0075)12t
0 1 2 3 4 5 6 7 8 t
Year
9. a. The initial value is $21,500, and the rate of decay is 9%,
or 0.09.
y = a(1 − r)t
y = 21,500(1 − 0.09)t
y = 21,500(0.91)t
The value of the car can be represented by
y = 21,500(0.91)t.
b. y = 21,500(0.91)t
= 21,500(0.91)(1/12)(12t)
= 21,500(0.911/12)12t
≈ 21,500(0.992)12t
So, 1 − r ≈ 0.992
−1
−1
−r ≈ −0.008
0.008
−r
— ≈ −—
−1
−1
r ≈ 0.008
The monthly percent decrease is about 0.8%.
So, the rate of growth is 20%.
6. f(t) = 3(1.02)10t
= 3(1.0210)t
≈ 3(1.22)t
So, the function represents exponential growth.
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Algebra 1
Worked-Out Solutions
327
Chapter 6
c.
t
y
0
4
8
21,500 14,744 10,110
12
6933
16
4755
7.
Value (dollars)
The initial amount is a = 25, and the rate of growth is
r = 0.2, or 20%.
y = 25(1.2)t
y = 21,500(0.91)t
y = 25(1.2)5
≈ 25(2.488)
≈ 62.2
So, the value of y is about 62.2 when t = 5.
0 2 4 6 8 10 12 14 16 t
Year
8.
r = 0.05
y = 12(1.05)t
Vocabulary and Core Concept Check
y = 12(1.05)5
1. In the exponential growth function y = a(1 + r)t, the
≈ 12(1.276)
quantity r is called the rate of growth.
≈ 15.3
2. The decay factor is 1 − r.
So, the value of y is about 15.3 when t = 5.
3. Exponential growth occurs when a quantity increases by the
9.
1 + r = 1.074
−1
The initial amount is a = 1500, and the rate of growth is
r = 0.074, or 7.4%.
b > 1 and x represents time. The function y = abx represents
exponential decay when 0 < b < 1 and x represents time.
f (t) = 1500(1.074)t
Monitoring Progress and Modeling with Mathematics
f (5) = 1500(1.074)5
5. The initial amount is a = 350, and the rate of growth is
≈ 1500(1.429)
r = 0.75, or 75%.
≈ 2143.4
0.75)t
So, the value of f(t) is about 2143.4 when t = 5.
y = 350(1.75)5
≈ 350(16.413)
≈ 5744.6
So, the value of y is about 5744.6 when t = 5.
6. The initial amount is a = 10, and the rate of growth is
r = 0.4, or 40%.
y = 10(1 + 0.4)t
y = 10(1.4)5
≈ 10(5.378)
≈ 53.8
So, the value of y is about 53.8 when t = 5.
Algebra 1
Worked-Out Solutions
−1
r = 0.074
4. The function y = abx represents exponential growth when
y = 350(1 +
−1
The initial amount is a = 12, and the rate of growth is
r = 0.05, or 5%.
6.4 Exercises (pp. 319–322)
same factor over equal intervals of time. Exponential decay
occurs when a quantity decreases by the same factor over
equal intervals of time.
1 + r = 1.05
−1
From the graph, you can see that the y-value is about 7000
when t = 12. So, the value of the car is about $7000 after
12 years.
328
−1
r = 0.2
Value of a Car
y
24,000
21,000
18,000
15,000
12,000
9000
6000
3000
0
1 + r = 1.2
−1
10.
1 + r = 1.028
−1
−1
r = 0.028
The initial amount is a = 175, and the rate of growth is
r = 0.028, or 2.8%.
h(t) = 175(1.028)t
h(5) = 175(1.028)5
≈ 175(1.148)
≈ 200.9
So, the value of h(t) is about 200.9 when t = 5.
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Chapter 6
11.
17. a. The initial amount is 315,000, and the rate of growth is
1+r=2
−1
2%, or 0.02.
−1
y = a(1 + r)t
r=1
y = 315,000(1 + 0.02)t
The initial amount is a = 6.72, and the rate of growth is
r = 1, or 100%.
y = 315,000(1.02)t
The population of Brookfield can be represented by
y = 315,000(1.02)t.
g(t) = 6.72(2)t
g(5) =
6.72(2)5
b. The value t = 20 represents 2020 because t = 0 represents
= 6.72(32)
2000.
= 215.0
y = 315,000(1.02)t
So, the value of g(t) is about 215.0 when t = 5.
12.
y = 315,000(1.02)20
≈ 315,000(1.4859)
1 + r = 1.8
−1
≈ 468,073
−1
r = 0.8
So, in 2020, the population will be about 468,000.
The initial amount is a = 1, and the rate of growth is
r = 0.8, or 80%.
18. a. The initial amount is 0.1, and the rate of growth is 23%,
or 0.23.
p(t) = 1.8t
y = a(1 + r)t
p(5) = 1.85
y = 0.1(1 + 0.23)t
≈ 18.9
y = 0.1(1.23)t
So, the value of p(t) is about 18.9 when t = 5.
The weight of the catfish during the 8-week period can be
represented by y = 0.1(1.23)t.
13. The initial amount is 10,000, and the rate of growth is 65%,
or 0.65.
b. y = 0.1(1.23)t
y = a(1 + r)t
y = 10,000(1 +
y = 0.1(1.23)4
0.65)t
≈ 0.1(2.289)
y = 10,000(1.65)t
≈ 0.229
The sales can be represented by y = 10,000(1.65)t.
So, after 4 weeks, the catfish will weigh about
0.229 pound.
14. The initial amount is 35,000, and the rate of growth is 4%,
or 0.04.
y = a(1 + r)t
y = 35,000(1 + 0.04)t
y = 35,000(1.04)t
Your salary can be represented by y = 35,000(1.04)t.
15. The initial amount is 210,000, and the rate of growth is
12.5%, or 0.125.
y = a(1 + r)t
y = 210,000(1 + 0.125)t
y = 210,000(1.125)t
The population can be represented by y = 210,000(1.125)t.
16. The initial amount is 4.5, and the rate of growth is 3.5%,
or 0.035.
19.
1 − r = 1 − 0.6
−1
−1
−r = −0.6
−r −0.6
—=—
−1
−1
r = 0.6
The initial amount is a = 575, and the rate of decay is
r = 0.6, or 60%.
y = 575(1 − 0.6)t
y = 575(0.4)3
= 575(0.064)
= 36.8
So, the value of y is 36.8 when t = 3.
y = a(1 + r)t
y = 4.5(1 + 0.035)t
y = 4.5(1.035)t
The cost of the item can be represented by y = 4.5(1.035)t.
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Algebra 1
Worked-Out Solutions
329
Chapter 6
20.
1 − r = 1 − 0.15
−1
23.
−1
1−r=
−1
−r = −0.15
−r = −0.005
−r −0.005
—=—
−1
−1
r = 0.005
−r = −0.15
−1
−1
r = 0.15
The initial amount is a = 700, and the rate of decay is
r = 0.005, or 0.5%.
The initial amount is a = 8, and the rate of decay is
r = 0.15, or 15%.
w(t) = 700(0.995)t
y = 8(1 − 0.15)t
w(3) = 700(0.995)3
y = 8(0.85)3
≈ 700(0.9851)
≈ 8(0.614)
≈ 689.6
≈ 4.9
So, the value of w(t) is about 689.6 when t = 3.
So, the value of y is about 4.9 when t = 3.
21.
1−r=
−1
24.
0.75
−1
The initial amount is a = 1250, and the rate of decay is
r = 0.135, or 13.5%.
h(t) = 1250(0.865)t
g(t) = 240(0.75)t
h(3) = 1250(0.865)3
g(3) = 240(0.75)3
≈ 1250(0.6472)
≈ 240(0.422)
≈ 809.0
≈ 101.3
So, the value of h(t) is about 809.0 when t = 3.
So, the value of g(t) is about 101.3 when t = 3.
−1
0.5
− 1.0
−r = −0.5
−r −0.5
—=—
−1
−1
r = 0.5
The initial amount is a = 475, and the rate of decay is
r = 0.5, or 50%.
f (t) = 475(0.5)t
f(3) = 475(0.5)3
= 475(0.125)
≈ 59.4
So, the value of f(t) is about 59.4 when t = 3.
− 1.000
−r = −0.135
−r −0.135
—=—
−1
−1
r = 0.135
The initial amount is a = 240, and the rate of decay is
r = 0.25, or 25%.
1−r=
1 − r = 0.865
−1
−r = −0.25
−r −0.25
—=—
−1
−1
r = 0.25
22.
0.995
− 1.000
25.
7
1−r=—
8
−1
−1
⋅
1
−r = −—
8
( )
(
7
8
7
8
8
8
1
8
— − 1 = — − — = −—
)
1
−1 (−r) = −1 −—
8
1
r = —, or 0.125
8
The initial amount is a = 1, and the rate of decay is
r = 0.125, or 12.5%.
()
()
7 t
y= —
8
7 3
y= —
8
≈ 0.7
So, the value of y is about 0.7 when t = 3.
330
Algebra 1
Worked-Out Solutions
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Chapter 6
26.
3
1−r= —
4
−1
−1
1
−r = −—
4
−1
31. Because the rate of growth is 150%, or 1.5, the growth factor
is 1 + 1.5 = 2.5, not just 1.5.
(
( )
3 4
1
3
— − 1 = — − — = −—
4 4
4
4
y = a(1 + r)t
)
b(t) = 10(1 + 1.5)t
b(8) = 10(2.5)8
−—
⋅ (−r) = −1
4
1
1
r = —, or 0.25
4
The initial amount is a = 0.5, and the rate of decay is
r = 0.25, or 25%.
3 t
y = 0.5 —
4
3 3
y = 0.5 —
4
≈ 0.5(0.422)
()
()
≈ 10(1525.879)
≈ 15,259
After 8 hours, there are about 15,259 bacteria in the culture.
32. Because 14% is the rate of decay, the decay factor is
1 − 0.14 not 1 + 0.14.
y = a(1 − r)t
v(t) = 25,000(1 − 0.14)t
v(5) = 25,000(0.86)5
≈ 0.2
≈ 25,000(0.4704)
So, the value of y is about 0.2 when t = 3.
≈ 11,761
27. The initial amount is 100,000, and the rate of decay is
The value of the car in 2015 is about $12,000.
2%, or 0.02.
y = 100,000(1 −
y=
0.02)t
100,000(0.98)t
The population can be represented by y = 100,000(0.98)t.
x
−1
0
1
2
y
50
10
2
0.4
9%, or 0.09.
y = a(1 − r)t
represents an exponential decay function.
0.09)t
⤻
+1 ⤻
+1 ⤻
+1
34.
y = 900(0.91)t
The cost of the sound system can be represented by
y = 900(0.91)t.
x
0
1
2
3
y
32
28
24
20
As x increases by 1, y decreases by 4. The function has
a constant rate of change. So, it is a linear function and
therefore neither an exponential growth nor an exponential
decay function.
9.5%, or 0.095.
r)t
y = 100(1 − 0.095)t
y = 100(0.905)t
The value of the stock can be represented by y = 100(0.905)t.
+1 ⤻
+1 ⤻
+1
⤻
35.
30. The initial amount is 20,000, and the rate of decay
is 13.4%, or 0.134.
y = a(1 − r)t
x
0
1
2
3
y
35
29
23
17
⤻ ⤻⤻
+(−6) +(−6) +(−6)
As x increases by 1, y decreases by 6. The function has
a constant rate of change. So, it is a linear function and
therefore neither an exponential growth nor an exponential
decay function.
y = 20,000(1 − 0.134)t
y = 20,000(0.866)t
The company’s profit can be represented by
y = 20,000(0.866)t.
⤻ ⤻⤻
+(−4) +(−4) +(−4)
29. The initial amount is 100, and the rate of decay is
y = a(1 −
⤻
⤻
⤻
1
1
1
×— ×— ×—
5
5
5
As x increases by 1, y is multiplied by —15 . So, the table
28. The initial amount is 900, and the rate of decay is
y = 900(1 −
⤻
+1 ⤻
+1 ⤻
+1
33.
y = a(1 − r)t
+1 ⤻
+1 ⤻
+1
⤻
36.
x
1
2
3
4
y
17
51
153
459
⤻⤻
⤻
×3 ×3 ×3
As x increases by 1, y is multiplied by 3. So, the table
represents an exponential growth function.
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Algebra 1
Worked-Out Solutions
331
Chapter 6
+5 ⤻
+5 ⤻
+5
⤻
37.
42. The function is of the form y = a(1 + r)t, where 1 + r > 1.
x
5
10
15
20
y
2
8
32
128
So, it represents exponential growth. Use the growth factor
1 + r to find the rate of growth.
1 + r = 1.1
⤻⤻
⤻
×4 ×4 ×4
−1
As x increases by 5, y is multiplied by 4. So, the table
represents an exponential growth function.
So, the rate of growth is 10%.
+2 ⤻
+2 ⤻
+2
⤻
38.
43. The function is of the form y = a(1 − r)t, where 1 − r < 1.
x
3
5
7
9
y
432
72
12
2
So, it represents exponential decay. Use the decay factor
1 − r to find the rate of decay.
⤻
⤻
⤻
1
1
1
1−r=
−1
×— ×— ×—
6
6
6
As x increases by 2, y is multiplied by —16. So, the table
+1
+1
Value
$37,000
2
$29,600
3
$23,680
4
$18,944
44. The function is of the form y = a(1 + r)t, where 1 + r > 1.
×0.8
So, it represents exponential growth. Use the growth factor
1 + r to find the rate of growth.
×0.8
×0.8
1 + r = 1.08
As t increases by 1, the value is multiplied by 0.8. So, the
table represents an exponential decay function.
b. (18,944)(0.8) = 15,155.2
+1
+1
+1
⤻
⤻
⤻
t
42
43
44
45
45
46
47
Visitors 11,000 12,100 13,310 14,641 16,105 17,716
⤻
⤻
×1.1
×1.1
So, after the website is online 47 days, about
17,716 people will have visited it.
41. The function is of the form y = a(1 − r)t, where 1 − r < 1.
So, it represents exponential decay. Use the decay factor
1 − r to find the rate of decay.
1 − r = 0.8
−1
−1
−r = −0.2
−0.2
−r —
—=
−1
−1
r = 0.2
So, the rate of decay is 20%.
332
Algebra 1
Worked-Out Solutions
−1
r = 0.06
⤻⤻
44
45. The function is of the form y = a(1 + r)t, where 1 + r > 1.
1 + r = 1.06
As t increases by 1, the number of visitors is multiplied
by 1.1. So, the table represents an exponential growth
function.
+1
+1
b.
43
r = 0.08
−1
⤻
⤻
⤻
×1.1
×1.1
×1.1
42
−1
So, it represents exponential growth. Use the growth factor
1 + r to find the rate of growth.
Visitors 11,000 12,100 13,310 14,641
t
−1
So, the rate of growth is 8%.
So, after 5 years, the value of the camper is about $15,155.
40. a.
−1
So, the rate of decay is 5%.
⤻⤻⤻
⤻⤻⤻
+1
t
1
0.95
−r = −0.05
−0.05
−r —
—=
−1
−1
r = 0.05
represents an exponential decay function.
39. a.
−1
r = 0.1
So, the rate of growth is 6%.
46. The function is of the form y = a(1 − r)t, where 1 − r < 1.
So, it represents exponential decay. Use the decay factor
1 − r to find the rate of decay.
1 − r = 0.48
−1
−1
−r = − 0.52
−0.52
−r —
—=
−1
−1
r = 0.52
So, the rate of decay is 52%.
47. The function is of the form y = a(1 + r)t, where 1 + r > 1.
So, it represents exponential growth. Use the growth factor
1 + r to find the rate of growth.
5
1+r=—
4
−1
−1
(
5 4 1
1
5
r = —, or 0.25
—−1=—−—=—
4
4 4 4
4
So, the rate of growth is 25%.
)
Copyright © Big Ideas Learning, LLC
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Chapter 6
48. The function is of the form y = a(1 − r)t, where 1 − r < 1.
So, it represents exponential decay. Use the decay factor
1 − r to find the rate of decay.
4
1−r=—
5
−1
−1
1
−r = −—, or −0.2
5
So, the rate of decay is 20%.
(
4
5
4
5
5
5
1
5
— − 1 = — − — = −—
)
54. f(t) = 0.4(1.16)t−1
(1.16)t
= 0.4—1
(1.16)
0.4
= —(1.16)t
1.16
≈ 0.34(1.16)t
The function is of the form y = a(1 + r)t, where 1 + r > 1.
So, it represents exponential growth.
55. b(t) = 4(0.55)t+3
49. y = (0.9)t−4
⋅ (0.55)
=
⋅ (0.55)
= 4(0.166375) ⋅ (0.55)
(0.9)t
= —4
(0.9)
(0.9)t
=—
0.6561
1
= — (0.9)t
0.6561
≈ 1.52(0.9)t
3
4(0.55)3
t
t
≈
⋅
The function is of the form y = a(1 − r)t, where 1 − r < 1.
So, it represents exponential decay.
0.67(0.55)t
The function is of the form y = a(1 − r)t, where 1 − r < 1.
So, it represents exponential decay.
56. r(t) = (0.88)4t
= (0.884)t
50. y = (1.4)t+8
≈ (0.60)t
⋅ (1.4)
≈ (1.4) ⋅ (14.8)
y = (1.4)t
= 4(0.55)t
8
The function is of the form y = a(1 − r)t, where 1 − r < 1.
So, it represents exponential decay.
t
≈ 14.8(1.4)t
The function is of the form y = a(1 + r)t, where 1 + r > 1.
So, it represents exponential growth.
51. y = 2(1.06)9t
= 2(1.069)t
r
n
(
)
nt
)
0.05 4t
y = 2000 1 + —
4
y = 2000(1 + 0.0125)4t
y = 2000(1.0125)4t
≈ 2(1.69)t
The function is of the form y = a(1 + r)t, where 1 + r > 1.
So, it represents exponential growth.
52. y = 5(0.82)t∕5
= 5(0.82)(1∕5)⋅t
(
r
n
58. y = P 1 + —
(
)
nt
)
0.10 2t
y = 1400 1 + —
2
y = 1400(1 + 0.05)2t
y = 1400(1.05)2t
= 5(0.821∕5)t
(
r
n
59. y = P 1 + —
5— t
= 5( √0.82 )
≈ 5(0.96)t
The function is of the form y = a(1 −
So, it represents exponential decay.
(
57. y = P 1 + —
r)t,
where 1 − r < 1.
53. x(t) = (1.45)t∕2
= (1.45)(1∕2)⋅t
= (1.451∕2)t
— t
= ( √1.45 )
≈ (1.20)t
(
)
nt
)
0.084 12t
y = 6200 1 + —
12
y = 6200(1 + 0.007)12t
y = 6200(1.007)12t
(
r
n
60. y = P 1 + —
(
)
nt
)
0.092 4t
y = 3500 1 + —
4
y = 3500(1 + 0.023)4t
y = 3500(1.023)4t
The function is of the form y = a(1 + r)t, where 1 + r > 1.
So, it represents exponential growth.
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Algebra 1
Worked-Out Solutions
333
Chapter 6
+1 ⤻
+1 ⤻
+1 ⤻
+1
⤻
61. a. Tree A
Year, t
Basal area, A
0
1
120
132
2
3
y = P(1 + r)t
y = 300(1.08)t
4
145.2 159.7 175.7
⤻
⤻⤻⤻
×1.1 ×1.1 ×1.1 ×1.1
From the table, you know the initial basal area of Tree A
is 120 square inches, and it is multiplied by a growth
factor of 1.1 each year. So, P = 120 and 1 + r = 1.1.
y = P(1 + r)t
So, the function that represents the balance of the
investment account is y = 300(1.03)4t, and the function
that represents the balance of the savings account is
y = 300(1.08)t.
b. Investment account:
Year, t
0
1
2
3
Balance (dollars), y 300 337.65 380.03 427.73
y = 120(1.1)t
Accounts
y = P(1 + r)t
y = 154(1 + 0.06)t
y = 154(1.06)t
So, the function that represents the basal area of Tree A
after t years is y = 120(1.1)t and the basal area of Tree B
after t years is y = 154(1.06)t.
Balance (dollars)
The initial basal area of Tree B is 154 square inches, and
the rate of growth is 6%, or 0.06.
y
800
700
600
500
400
300
200
100
0
y = 300(1.03)4t
y = 300(1.08)t
0 1 2 3 4 5 6 7 8 t
b. Tree B
Year
Year, t
0
1
2
3
4
Basal area, A 154 163.24 173.03 183.42 194.42
Both accounts start with the same balance. The investment
account balance is increasing at a faster rate, so it is
greater than the savings account balance after the start.
Basal area (in.2)
Tree Basal Area
A
200
175
150
125
100
75
50
25
0
63. a. The initial value is 25,000, and the rate of growth is 5.5%,
or 0.055.
AB = 154(1.06)t
AA = 120(1.1)t
0 1 2 3 4 t
Year
The basal area of Tree B is larger than the basal area
of Tree A, but the difference between the basal areas is
decreasing.
y = P(1 + r)t
y = 25,000(1 + 0.055)t
y = 25,000(1.055)t
A function that represents the city’s population is
y = 25,000(1.055)t.
1
b. Use the fact that t = —(12t) and the properties of
12
exponents to rewrite the function in a form that reveals the
monthly rate of growth.
y = 25,000(1.055)t
y = 25,000(1.055)(1/12)(12t)
y = 25,000(1.055(1/12))(12t)
62. a. The principal of the investment account is $300, the
annual interest rate is 6%, or 0.06, and because the interest
is compounded quarterly, n = 4.
r
y=P 1+—
n
(
(
nt
)
y ≈ 25,000(1.00447)(12t)
1 + r ≈ 1.00447
−1
)
0.12 4t
y = 300 1 + —
4
y = 300(1 + 0.03)4t
−1
r ≈ 0.00447
So, the monthly percent increase is about 0.45%.
y = 300(1.03)4t
The graph crosses the y-axis at (0, 300). So, the principal
of the savings account is $300. The points on the graph
are approximately (1, 325), (2, 350), and (3, 375).
350
350
325
Because — ≈ 1.08, — ≈ 1.08, and — ≈ 1.07, the
300
325
325
balance of the savings account has a growth factor of
about 1.08. So, P = 300 and 1 + r = 1.08.
334
Algebra 1
Worked-Out Solutions
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Chapter 6
c.
Year, t
0
2
4
6
8
Population, y 25,000 27,826 30,971 34,471 38,367
65. a. Equation 1: y ≈ 800(0.71)t
Equation 2: y ≈ 800(0.9943)60t
≈ 800(0.994360)t
≈ 800(0.7097)t
Population
City Population
y
40,000
35,000
30,000
25,000
20,000
15,000
10,000
5000
0
Equation 3: y ≈ 800(0.843)2t
≈ 800(0.8432)t
≈ 800(0.7106)t
y = 25,000(1.055)t
So, all three equations are approximately equivalent to
y ≈ 800(0.71)t.
b. Equation 1:
a = 800
0 1 2 3 4 5 6 7 8 t
1−r=
−1
Year
−r
−1
y = a(0.5)t/x
r = 0.29
Equation 2:
y = 3(0.5)t/88
A function that represents the amount of plutonium–238 is
y = 3(0.5)t/88.
a = 800
1−r=
−1
−r
−1
Equation 3:
a = 800
–1
0.843
−1
−r = −0.157
–r –0.0078
–1
–1
r ≈ 0.0078
—≈—
−r
−1
−0.157
−1
—=—
r = 0.157
So, the yearly percent decrease is about 0.8%.
Time (years), t
0
Amount (grams), y
3 2.907 2.817 2.7294 2.6445
4
8
12
16
Plutonium-238 Decay
Amount (grams)
1−r=
−1
–r ≈ –0.0078
y
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0
−0.0057
−1
r = 0.0057
1 – r ≈ 0.9921
c.
−1
—=—
3(0.5(1/88))t
≈ 3(0.9921)t
–1
0.9943
−r = −0.0057
3(0.5)t/88
y = 3(0.5)(1/88)t
=
−0.29
−1
—=—
64. a. The initial amount is 3 grams, and the half-life x is 88.
y=
−1
−r = −0.29
From the graph, you can see that the population after
4 years is about 30,971.
b.
0.71
All three functions indicate the initial amount of ibuprofen
in a person’s bloodstream is 800 milligrams. The first
function indicates the amount of ibuprofen in a person’s
bloodstream decreases by about 29% each hour. The second
function indicates the amount of ibuprofen in a person’s
bloodstream decreases by about 0.57% each minute.
The third function indicates the amount of ibuprofen in
a person’s bloodstream decreases by about 15.7% each
half-hour.
y = 3(0.5)t/88
0 2 4 6 8 10 12 14 16 t
Time (years)
From the graph, you can see that the amount of
plutonium−238 remaining after 12 years is about 2.7 grams.
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Algebra 1
Worked-Out Solutions
335
Chapter 6
r
b(t) = P 1 + —
n
(
66. Savings account:
)
nt
72. a. The function f is an exponential growth function because as
(
0.036
b(t) = 9000 1 + —
12
t increases by 1, y is multiplied by 2. The graph crosses the
y-axis at (0, 1). So, the initial amount is 1. Because the growth
factor 1 + r is 2, the growth rate r is 2 − 1 = 1, or 100%.
)
12t
= 9000(1 + 0.003)12t
b. Because f (1) = 2 and g(1) = 8, the graph of g is a vertical
8
stretch of the graph of f by a factor of — = 4. So, k = 4.
2
c. g(t) = k f (t)
= 9000(1.003)12t
h(t) = 40(12)t
Safe at home:
h(t) = 480t
=4
C(t) = b(t) + h(t)
=
C(t) =
9000(1.003)12t
+ 480t
t
= f (t + 2)
amount of money saved both in the savings account and at home.
67. The growth factor is 3.
So, h(t) must be f (t + 2), which means that r = 2.
Maintaining Mathematical Proficiency
1+r=3
73.
−1
8x + 12 =
− 8x
r = 2, or 200%
68. a. D; The graph is increasing, and the value from x = 0 to
x = 2 increases from 40 to 160, which is the expected value.
b. B; The graph is decreasing, and the value from x = 0
− 8x
74. 5 − t = 7t + 21
c. A; The graph is increasing, and the value from x = 0 to
5 = 8t + 21
−21
−16
8
to x = 2 decreases from 500 to about 450, which is the
expected value.
70. Sample answer: Account A could earn 4.5% annual interest
compounded semiannually, and Account B could earn 6%
annual interest compounded monthly.
r nt
r nt
f (t) = P 1 + —
g(t) = P 1 + —
n
n
0.045 2t
0.06 12t
f (t) = 1000 1 + —
g(t) = 1000 1 + —
2
12
)
(
)
= 1000(1 + 0.0225)2t
= 1000(1 + 0.005)12t
=
= 1000(1.005)12t
1000(1.0225)2t
So, a function that represents the balance of Account A is
f (t) = 1000(1.0225)2t, and a function that represents the
balance of Account B is g(t) = 1000(1.005)12t. I would rather
use Account B because it has a higher annual interest rate
and it is compounded more frequently. So, it will grow at a
faster rate.
71. no; The discount is 20% of the preceding day’s price, not
always the original price, so the amount of the discount is
less each day.
Algebra 1
Worked-Out Solutions
7=7✓
8t
8
−2 = t
The solution is t = −2.
constant function equivalent to y = 5.
(
−21
—=—
69. Sample answer: y = 5(1)x; One to any power is 1, so this is a
)
5 − t = 7t + 21
?
5 − (−2) = 7(−2) + 21
?
5 + 2 = −14 + 21
−16 = 8t
d. C; The graph is decreasing, and the value from x = 0
(
−12 = −12 ✓
Check:
+t +t
x = 2 increases from 40 to about 50, which is the
expected value.
)
8x + 12 = 4x
?
8(−3) + 12 = 4(−3)
?
−24 + 12 = −12
The solution is x = −3.
to x = 2 decreases from 500 to about 350, which is the
expected value.
(
Check:
4x
12 = −4x
−4x
12
—=—
−4
−4
−3 = x
So, the growth rate is 200%.
336
t
= 2t+2
The function C(t) = 9000(1.003)12t + 480t represents the total
−1
⋅ (2)
⋅ (2)
22
75.
6(r − 2) = 2r + 8
Check:
6(r) − 6(2) = 2r + 8
6r − 12 = 2r + 8
−2r
−2r
4r − 12 =
+12
6(r − 2) = 2r + 8
?
6(5 − 2) = 2(5) + 8
?
6(3) = 10 + 8
18 = 18 ✓
8
+12
4r = 20
4r
4
20
4
—=—
r=5
The solution is r = 5.
76. Because the equation y = −6x + 7 is in slope-intercept
form, the slope is −6, and the y-intercept is 7.
1
4
1
the slope is —, and the y-intercept is 7.
4
77. Because the equation y = —x + 7 is in slope-intercept form,
Copyright © Big Ideas Learning, LLC
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Chapter 6
— 5
8. ( √ 4 ) = ( √ 2
6x − 12
3
= (2)5
3y
3
—=—
= 32
y = 2x − 4
9.
Because the equation y = 2x − 4 is in slope-intercept form,
the slope is 2, and the y-intercept is −4.
2y + x = 8
79.
⋅2)
— 5
78. 3y = 6x − 12
x
−2
−1
0
1
2
5x
5−2
5−1
50
51
52
y
—
1
25
—
1
5
1
5
25
2y + x − x = 8 − x
2y = −x + 8
−x + 8
2
1
y = −—x + 4
2
2y
2
y=5
—=—
The domain is all real numbers
and the range is y > 0.
y
400
x
300
200
1
Because the equation y = −—x + 4 is in slope-intercept
2
1
form, the slope is −—, and the y-intercept is 4.
2
100
−4
−2
4 x
2
6.1–6.4 What Did You Learn? (p. 323)
1. Sample answer: Apply the distance formula (d = rt) to find
an equation solved for time, and apply the laws of exponents
to simplify the powers of 10.
10.
−2
x
x
()
1
−2 —
6
2. Sample answer: Apply what you previously established
about the sign of a radical with an odd index, which is the
same as the sign of the radicand.
−4
−2
y = −2( 6)
6.1–6.4 Quiz (p. 324)
( )
3s
(
) (
2
⋅ ⋅
⋅
1 1 x2
= —
2 y4
⋅
x2 2
⋅
43r2 3
64r6
) ( )
2
x2
= —4
2y
2
(x2)2
=—
(2y4)2
x
0
6(2)x−4 −1
⋅
5. √ 27 = √ 3 ⋅ 3 ⋅ 3 = 3
( 161 )
6. —
1/4
1
−—
3
()
1
−2 —
6
0.5
1
−—
18
The domain is all real
numbers and the range
is y < 0.
y = 6(2)x−4 − 1
⋅ ⋅ ⋅
⋅ ⋅ ⋅
4 ——
7. 5122/3 = (5121/3)2
4
6
8
5
23
95
The domain is all real
numbers and the range
is y > −1.
60
√1
√1 1 1 1 1
11/4
= ——
=—
=—
=—
4—
4 ——
161/4 √
2
16 √ 2 2 2 2
40
20
2
4
6
8 x
3— 2
= ( √ 512 )
⋅⋅
3— 2
= ( √8 8 8 )
= (8)2
2
6(2)−4 −1 6(2)−2 −1 6(2)0 −1 6(2)2 −1 6(2)4 −1
y
3—
4—
2
−0.625
y
x4
=—
=—
22 y4⋅2 4y8
3—
1
11.
43(r2)3
=—
=—=—=—
(3s5)3 33(s5)3 33s5⋅3 27s15
2x0
4x y
4 x
()
1
−2 —
6
−400
(4r2)3
4. —
−2 4
−2
0
−300
k−12
1
=—
k12
3. —5
−12
2
()
2
−200
2. (k4)−3 = k4(−3)
4r2 3
1
−2 —
6
1
−100
= 32+4 = 36
=
−1
y
1 x
4
0
()
1
−2 —
6
−72
y
exponential function, and the second term, h(t), is in the form
of a linear function.
⋅3
−2
()
1
−2 —
6
3. Sample answer: The first term, b(t), is in the form of an
1. 32
−1
= 64
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Algebra 1
Worked-Out Solutions
337
Chapter 6
+1
⤻
12.
+1
⤻
+1
⤻
103
10
17. a. —
= 103−(−9)
−9
x
0
1
2
3
= 103+9
y
7
21
63
189
= 1012
⤻
⤻
×3
×3
⤻
A kilogram is 1012, or 1,000,000,000,000, times larger
than a nanogram.
×3
As x increases by 1, y is multiplied by 3. So, the table
represents an exponential growth function.
+1
⤻
13.
+1
⤻
+1
⤻
1
2
3
4
y
14,641
1331
121
11
⤻
1
= 104−3 g = 101 g = 10 g
1000 dg
1
⤻
1
×—
11
×—
11
—
×—
11
1
As x increases by 1, y is multiplied by —. So, the table
11
represents an exponential decay function.
14. The function is of the form y = a(1 + r)t, where 1 + r > 1.
So, it represents exponential growth. Use the growth factor
1 + r to find the rate of growth.
−1
g
18. V = lwh
6—
V = (1634)(√ 64 )(24315)
⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 )(√243 )
= [ (√ 16 ) ](2)(√3 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 3 )
= [(√2 ⋅ 2 ⋅ 2 ⋅ 2 ) ](2)(3)
6 ——
= [ (1614)3 ](√ 2
4— 3
5—
5 ——
4 —— 3
= (23)(2)(3)
r = 0.88
= (8)(2)(3)
So, the rate of growth is 88%.
= 16(3)
15. The function is of the form y = a(1 + r)t, where 1 + r > 1.
So, it represents exponential growth. Use the growth factor
1 + r to find the rate of growth.
= 48 ft3
The volume of the cedar chest is 48 cubic feet.
19. a. The function f (t) = 5(4)t is of the form y = a(1 + r)t,
1 + r = 1.26
where 1 + r > 1. So, it represents exponential growth.
−1
b.
r = 0.26
So, the rate of growth is 26%.
16. The function is of the form y = a(1 − r)t, where 1−r < 1.
t
0
1
2
3
4
5(4)t
5(4)0
5(4)1
5(4)2
5(4)3
5(4)4
f (t)
5
20
80
320
1280
Frog Pond
3
1−r=—
5
−1
−1
(
2
−r = −—
5
3
5
3
5
5
5
2
5
— − 1 =— − — = −—
⋅ (−r) = −1( −—5 )
2
2
r = —, or 0.4
5
So, the rate of decay is 40%.
)
Frogs
So, it represents exponential decay. Use the decay factor 1−r
to find the rate of decay.
−1
3
So, 1000 decigrams is greater because 1000 decigrams is
100 grams, and 10,000 milligrams is 10 grams.
−1
−1
−1
10 g
= 10 ⋅ 10
⋅—
1 dg
= 103−1 g = 102 g = 100 g
1 + r = 1.88
−1
⋅
⋅
x
⤻
1
102
10
A milligram is 105, or 100,000, times smaller than a
hectogram.
10,000 mg 10−3 g 104 10−3 g
c. — — = ——
1
1 mg
1
b. —
= 102−(−3) = 102+3 = 105
−3
y
2400
2100
1800
1500
1200
900
600
300
0
f(t) = 5(4)t
0 1 2 3 4 5 6 7 8 t
Year
Based on the context of the problem, you cannot have a
negative amount of time pass. So, the domain is t ≥ 0.
Because f (0) = 5, and the graph increases after that, the
range is y ≥ 5.
c. Use the growth factor 1 + r to find the rate of growth.
338
Algebra 1
Worked-Out Solutions
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Chapter 6
1+r=4
1
3. a. Graph y = 2x and y = —2 .
−1 −1
3
r=3
So, the yearly percent change is r = 3, or 300%.
f (t) = 5(4)t
=
−3
5(4)(112)(12t)
= 5(4112)12t
≈ 5(1.1225)12t
1 + r ≈ 1.1225
−1
3
Intersection
X=-1
Y=.5
−1
The solution is x = −1.
b. Graph y = 2x+1 and y = 0.
6
−1
r ≈ 0.1225
So, the monthly percent change is about r = 0.1225, or
12.25%.
d. f (t) = 5(4)t
−6
6
−2
The graph of y = 2x+1 does not intersect the line y = 0,
which is the x-axis. So, the equation has no solution.
f (4) = 5(4)4
—
= 5(256)
c. Graph y = 2x and y = √ 2 .
= 1280
3
So, after 4 years, there are 1280 frogs in the pond.
6.5 Explorations (p. 325)
1. Graph y = 2.5x−3 and y = 6.25. Use the intersect feature of
the graphing calculator to find the coordinates of the point of
intersection.
−3
Intersection
X=.5
Y=1.4142136
−1
3
The solution is x = 0.5, or —12.
d. Graph y = 3x and y = 9.
7
12
−6 Intersection
X=5
Y=6.25
−1
6
The x-coordinate of the point of intersection, 5, is the
solution of the equation. So, the solution is x = 5.
2. a– c. Sample answer:
−12
12
Intersection
X=2
Y=9
−4
The solution is x = 2.
e. Graph y = 3x−1 and y = 0.
6
7
−6
4
−6
6
−2
−3
The graphs shown are y = 2x, y = −1, and y = x + 3.
d. Yes, it is possible for an exponential equation to have no
solution or more than one solution. The graphs from parts
(a) – (c) show that 2x = −1 has no solution and 2x = x + 3
has two solutions.
The graph of y = 3x−1 does not intersect the line y = 0,
which is the x-axis. So, the equation has no solution.
f. Graph y = 42x and y = 2.
3
−3
Intersection
X=.25
Y=2
−1
3
1
The solution is x = 0.25, or —.
4
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Algebra 1
Worked-Out Solutions
339
Chapter 6
1
4
g. Graph y = 2x2 and y = —.
3x + 5 = x + 1
3
−5
−x
−x
2x + 5 = 1
−5 −5
1
Intersection
X=-4
Y=.25
73x+5 = 7x+1
3.
2x = −4
2x −4
—=—
2
2
x = −2
−1
The solution is x = −4.
1
9
h. Graph y = 3x+2 and y = —.
Check: 73x+5 = 7x+1
?
73(−2)+5 = 7−2+1
?
7−6+5 = 7−1
? 1
7−1 = —
7
1 1
—=—✓
7 7
The solution is x = −2.
3
4. 4x = 256
−5
Intersection
X=-4
Y=.11111111
−1
4x
1
The solution is x = −4.
=
Check:
44
x=4
256 = 256 ✓
The solution is x = 4.
3
2
i. Graph y = 2x−2 and y = — x − 2.
5.
92x = 3x−6
Check:
(32)2x = 3x−6
6
32⋅2x = 3x−6
34x = 3x−6
−6
6
4x = x − 6
−2
−x
The solutions are x = 2 and x = 4.
original equation equal to y. Graph the new equations. The
x-coordinate(s) of the point(s) of intersection is/are the
solution(s).
5. Graph y = 30(2n) and y = 960.
The solution is x = −2.
6.
43x = 8x+1
22(3x) = 23(x+1)
43x = 8x+1
?
= 81+1
?
43 = 82
2(3x) = 3(x + 1)
64 = 64 ✓
(22)3x
1100
=
Check:
(23)x+1
43(1)
6x = 3(x) + 3(1)
6
6x = 3x + 3
The solution is n = 5. So, there will be 960 mice in the
population in 5 years.
6.5 Monitoring Progress (pp. 326–328)
1. 22x = 26
−x
92x = 3x−6
?
92(−2) = 3−2−6
?
9−4 = 3−8
1 ? 1
—4 = —8
9
3
1
1
—=—
6561 6561
3x = −6
3x −6
—=—
3
3
x = −2
4. Form two new equations by setting each side of the
Intersection
Y=960
0 X=5
0
4x = 256
?
44 = 256
Check:
2x = 6
2x 6
—=—
2
2
x=3
22x = 26
?
22⋅3 = 64
?
26 = 64
− 3x − 3x
3x = 3
3x 3
—=—
3
3
x=1
The solution is x = 1.
64 = 64 ✓
The solution is x = 3.
2. 52x = 5x+1
−x−x
52x = 5x+1
?
52(1) = 51+1
?
52 = 52
x=1
25 = 25 ✓
Check:
2x = x + 1
The solution is x = 1.
340
Algebra 1
Worked-Out Solutions
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Chapter 6
( 13 )
—
7.
x−1
( 13 )
( 13 )
( 13 )
= 27
—
Check:
(3−1)x−1 = 33
—
x−1
−2−1
= 27
?
= 27
3. 45x = 410
?
= 27
—
?
33 = 27
−1(x − 1) = 3
−(x) − 1(−1) = 3
−1
+4
x = 12
5.
5,764,801 = 5,764,801 ✓
39x = 37x+8
Check:
9x = 7x + 8
2x = 8
2x 8
—=—
2
2
x=4
3
336 = 336
1.50095 × 1017 = 1.50095 × 1017 ✓
The solution is x = 4.
The solution is x ≈ 0.85.
6. 24x = 2x+9
9. Graph y = 4x−3 and y = x + 2.
Check:
4x = x + 9
−x−x
7
−6
3x = 9
3x 9
—=—
3
3
x=3
6
−1
The solutions are x ≈ −2 and x ≈ 4.33.
()
39x = 37x+8
?
= 37(4)+8
?
336 = 328+8
39(4)
− 7x − 7x
3
1
4
+4
7x−4 = 78
?
712−4 = 78
?
78 = 5,764,801
The solution is x = 12.
8. Graph y = 2x and y = 1.8.
10. Graph y = —
Check:
x−4=8
The solution is x = −2.
Intersection
X=.84799691 Y=1.8
−1
1,048,576 = 1,048,576 ✓
4. 7x−4 = 78
−x = 2
2
−x
—=—
−1 −1
x = −2
−3
45x = 410
?
45⋅2 = 410
?
410 = 1,048,576
The solution is x = 2.
27 = 27 ✓
−x + 1 = 3
Check:
5x = 10
5x 10
—=—
5
5
x=2
−3
3−1(x−1) = 33
−1
Monitoring Progress and Modeling with Mathematics
x
and y = −2x − 3.
24x = 2x+9
?
24(3) = 23+9
?
212 = 212
4096 = 4096 ✓
The solution is x = 3.
7. 2x = 64
Check:
2x = 26
x=6
6
2x = 64
?
26 = 64
64 = 64 ✓
The solution is x = 6.
−6
6
8. 3x = 243
3x
−2
=
x=5
The graphs do not intersect. So, the equation has no solution.
2. The equation that does not belong with the other three is
34 = x + 42. Although the equation contains exponents, it is
not an exponential equation. The other three equations are
exponential because at least one of the exponents contains
a variable.
Copyright © Big Ideas Learning, LLC
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243 = 243 ✓
7x−5 = 49x
9.
7x−5 = (72)x
1. Rewrite each side of the equation using the same base,
then set the exponents equal to each other and solve. If it is
impossible to rewrite each side of an exponential equation
using the same base, you can solve the equation by graphing
each side and finding the point(s) of intersection.
3x = 243
?
35 = 243
The solution is x = 5.
6.5 Exercises (pp. 329–330)
Vocabulary and Core Concept Check
Check:
35
7x−5 = 72x
x − 5 = 2x
−x
−x
−5 = x
7x−5 = 49x
?
7−5−5 = 49−5
? 1
7−10 = —5
49
1 ?
1
=
—
——
710 282,475,249
1
1
—— = ——
282,475,249 282,475,249
Check:
The solution is x = −5.
Algebra 1
Worked-Out Solutions
341
Chapter 6
10. 216x = 6x+10
216x = 6x+10
?
2165 = 65+10
?
2165 = 615
Check:
(63)x = 6x+10
63x = 6x+10
3x = x + 10
−x
x
14.
(26)2x+4 = (24)5x
26(2x+4) = 24(5x)
6(2x + 4) = 4(5x)
6(2x) + 6(4) = 20x
Check: 642x+4 = 165x
?
642(3)+4 = 165(3)
?
646+4 = 1615
?
6410 = 1615
(33)x = (32)x−2
33x = 32(x−2)
3x = 2(x − 2)
3x = 2(x) − 2(2)
3x = 2x − 4
1.15292 × 1018 = 1.15292 × 1018 ✓
−7 = 5x + 3
−3
−10 5x
5
5
−2 = x
The solution is x = −2.
27x = 9x−2
?
27−4 = 9−4−2
1 ?
—4 = 9−6
27
1
1
— = —6
531,441 9
1
1
—=—
531,441 531,441
Check:
Check:
(5−1)x = 53
=
53
5−x
=
53
16.
1
34x−9 = —
243
1
34x−9 = —5
3
34x−9 = 3−5
4x − 9 = −5
+9 +9
4x = 4
The solution is x = −4.
5−1(x)
1
128
1 ?
— = 25(−2)+3
128
1 ?
— = 2−10+3
128
1 ?
— = 2−7
128
1 ? 1
— = —7
128 2
1
1
—=—✓
128 128
— = 25x+3
—=—
x = −4
()
−3
−10 = 5x
− 2x
= 125
Check:
2−7 = 25x+3
− 12x
27x = 9x−2
x
256 = 256 ✓
1
128
1
—7 = 25x+3
2
The solution is x = 3.
1
5
44 = 256
15. — = 25x+3
24 = 8x
24 8x
—=—
8
8
3=x
—
−4
The solution is x = −4.
12x + 24 = 20x
13.
—
−x = 4
4
−x
—=—
−1 −1
x = −4
642x+4 = 165x
− 2x
( 14 ) = 256
( 14 ) =?? 256
—
4−x = 44
The solution is x = 5.
12.
Check:
4−1(x) = 44
4.70185 × 1011 = 4.70185 × 1011
2x = 10
2x 10
—=—
2
2
x=5
− 12x
—
(4−1)x = 44
−x
11.
x
( 14 ) = 256
x
( ) = 125
( 15 ) =?? 125
1
5
—
—
−3
4x 4
—=—
4
4
x=1
Check:
1
34x−9 = —
243
? 1
34(1)−9 = —
243
? 1
4−9
3
=—
243
? 1
−5
3 =—
243
1 ? 1
—5 = —
3
243
1
1
—=—✓
243 243
The solution is x = 1.
53 = 125
125 = 125
−x = 3
3
−x
−1 −1
x = −3
—=—
The solution is x = −3.
342
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Chapter 6
17.
( )
( )
1 x+1
36−3x+3 = —
216
1 x+1
(62)−3x+3 = —3
6
62(−3x+3) = (6−3)x+1
Check:
2(−3x + 3) = −3(x + 1)
+3
+3
9 = 3x
9 3x
—=—
3
3
3=x
14
53x+2 = 25x−8
53x+2 = 52(x−8)
3x + 2 = 2(x − 8)
3x + 2 = 2x − 16
( )
( )
( )
1
2,176,782,336
—— = —— ✓
( )
( )
4–x
4–x
= 92x−1
= (32)2x−1
(3−3)4−x = 32(2x−1)
3−3(4−x) = 32(2x−1)
−3(4 −x) = 2(2x − 1)
−3(4) − 3(−x) = 2(2x) − 2(1)
−12 + 3x = 4x − 2
− 3x − 3x
− 12 = x − 2
+2
?
= 9−21
the exponents are set equal to each other.
3x + 2 = 2(x) − 2(8)
1
27
1
—3
3
?
= 9−20−1
19. The powers should be rewritten with the same base before
Check:
—
?
= 92(−10)−1
9.13918 × 10−21 = 9.13918 × 10−21 ✓
The solution is x = 3.
1
2,176,782,336
= 92x−1
53x+2 = (52)x−8
1 x+1
36−3x+3 = —
216
? 1 3+1
36−3(3)+3 = —
216
?
1 4
36−9+3 = —
216
? 1
−6
36 = —4
216
1 ?
1
—6 = ——
36
2,176,782,336
18.
4+10
—
−6x + 6 = −3x − 3
6 = 3x − 3
4–x
4–(–10)
—
2(−3x) + 2(3) = −3(x) − 3(1)
+ 6x
—
—
62(−3x+3) = 6−3(x+1)
+ 6x
( 271 )
( 271 )
( 271 )
( 271 )
+2
− 10 = x
The solution is x = −10.
− 2x
− 2x
x + 2 = −16
−2
−2
x = −18
The solution is x = −18.
1
2
1
8
20. The fraction — is equal to —3 = 2−3, not 23.
( 18 )
( 21 )
—
—3
5x
5x
= 32x+8
= (25)x+8
( 2−3 )5x = 25(x+8)
2−3(5x) = 25(x+8)
−3(5x) = 5(x + 8)
−15x = 5(x) + 5(8)
−15x = 5x + 40
− 5x
− 5x
−20x = 40
40
−20x
—=—
−20
−20
x = −2
The solution is x = −2.
21. C; The graph of y = 2x is a parent function with no
translations. It has a y-intercept of 1, and the y-values
increase as x increases because the base 2 is greater than 1.
So, the solution is x ≈ 2.58.
22. D; The graph of y = 42x−5 is a horizontal shrink and a
horizontal translation to the right of the graph of the parent
function y = 4x. Because the base 4 is greater than 1, the
y-values increase as x increases. So, the solution is x ≈ 3.15.
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Algebra 1
Worked-Out Solutions
343
Chapter 6
23. B; The graph of y = 5x+2 is a horizontal translation 2 units
30. Graph y = 3x − 2 and y = 5x−1
left of the graph of the parent function y = 5x. Because the
base 5 is greater than 1, the y-values increase as x increases.
So, the solution is x ≈ −0.89.
4
24. A; The graph of y = 3−x−1, or y = 3−(x+1), is a reflection
in the y-axis and a horizontal translation 1 unit left of the
graph of the parent function y = 3x. This is the only function
whose y-values decrease as x increases. So, the solution is
x ≈ −2.63.
25. Graph y =
6x+2
−3
3
0
The solutions are x = 1 and x ≈ 1.71.
( 13 )
1
2
31. Graph y = —x − 1 and y = —
and y = 12.
2x–1
.
5
14
−2
−12 Intersection
X=-.6131472 Y=12
−2
12
5
Intersection
X=2.0643126 Y=.03215632
−2
The solution is x ≈ 2.06.
The solution is x ≈ −0.61.
3
4
32. Graph y = 2−x+1 and y = −—x + 3.
26. Graph y = 5x−4 and y = 8.
6
12
−6
−12
12
Intersection
X=5.2920297 Y=8
−4
The solution is x ≈ 5.29.
()
7x+1
1
27. Graph y = —
2
and y = −9.
6
−2
The solutions are x ≈ −0.87 and x ≈ 3.81.
33. Graph y = 5x and y = −4−x+4.
4
10
−6
−5
6
5
−4
−10
The graphs do not intersect. So, the equation has no solution.
The graphs do not intersect. So, the equation has no solution.
()
1
28. Graph y = —
3
x+3
and y = 10.
34. Graph y = 7x−2 and y = 2−x.
3
12
−3
−12
Intersection
X=-5.095903 Y=10
−4
12
The solution is x ≈ −5.10.
29. Graph y = 2x+6 and y = 2x + 15.
Intersection
X=1.4747009 Y=.35980798
−1
3
The solution is x ≈ 1.47.
35. Graph y = 2−x−3 and y = 3x+1.
5
12
−12
12
−4
−5 Intersection
1
X=-1.773706 Y=.42741387
−1
The solution is x ≈ −1.77.
The solutions are x ≈ −7.30 and x ≈ −2.75
344
Algebra 1
Worked-Out Solutions
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Chapter 6
41.
36. Graph y = 5−2x+3 and y = −6x+5.
y = 2x+2
32 = 2x+2
8
25 = 2x+2
−12
5=x+2
12
−2
The graphs do not intersect. So, the equation has no solution.
⋅ 5 = 150
150
30 ⋅ 5
—=—
30
⋅
30 ⋅ 5
x+3
−2+3
30
=
−3
⋅
y = 192(4x−3)
42.
200,000 = 192(4x−3)
150 = 150 ✓
51
x+3=
?
= 150
?
30 51 = 150
5x+3 = 5
5x+3
The solution is x = 3. So, the photo must be enlarged
3 times on the computer so the new photo is 32 times the
original size.
Check: 30 5x+3 = 150
x+3
37. 30
−2
3=x
−8
Graph y = 192(4x−3) and y = 200,000,
1
250,000
−3
x=−2
The solution is x = −2.
⋅
⋅
⋅
2x−7 = 24
Check: 12 2x−7 = 24
24
12 2x−7 —
—=
12
12
?
12 ∙ 28−7 = 24
38. 12
Intersection
0 X=8.012339 Y=200000
0
?
12 ∙ 21 = 24
2x−7 = 2
2x−7 = 21
The solution is x ≈ 8.01. So, there will be 200,000 bacteria
after about 8 hours.
43.
33x+6 = 33(x+2)
3x + 6 = 3(x + 2)
+7
3x + 6 = 3(x) + 3(2)
x=8
The solution is x = 8.
39. 4(3−2x−4) = 36
4(3−2x−4) 36
4
4
3−2x−4 = 9
—=—
3−2x−4 = 32
−2x − 4 = 2
+4
3x + 6 = 3x + 6
Check: 4(3−2x−4) = 36
?
4(3−2(−3)−4) = 36
?
4(36−4) = 36
?
4(32) = 36
?
4(9) = 36
+4
− 3x
2(42x+1) 128
—=—
2
2
42x+1 = 64
42x+1 = 43
2x + 1 = 3
−1
− 3x
6=6
The equation 6 = 6 is always true. So, the equation has
infinitely many solutions.
44.
34x+3 = 81x
34x+3 = (34)x
36 = 36 ✓
34x+3 = 34x
−2x = 6
6
−2x —
—=
−2
−2
x = −3
The solution is x = −3.
40. 2(42x+1) = 128
33x+6 = 27x+2
33x+6 = (33)x+2
24 = 24 ✓
x−7=1
+7
10
4x + 3 = 4x
− 4x
− 4x
3=0✗
Check: 2(42x+1) = 128
?
2(42(1)+1) = 128
?
2(42+1) = 128
?
2(43) = 128
?
2(64) = 128
−1
2x = 2
2
2x —
—=
2
2
x=1
The solution is x = 1.
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128 = 128 ✓
The equation 3 = 0 is never true. So, the equation has
no solution.
4x+3 = 22(x+1)
45.
(22)x+3 = 22(x+1)
22(x+3) = 22(x+1)
2(x + 3) = 2(x + 1)
2(x) + 2(3) = 2(x) + 2(1)
2x + 6 = 2x + 2
− 2x
− 2x
6=2✗
The equation 6 = 2 is never true. So, the equation has
no solution.
Algebra 1
Worked-Out Solutions
345
Chapter 6
58(x−1) = 6252x−2
46.
58(x−1) = (54)2x−2
58(x−1) = 54(2x−2)
8(x − 1) = 4(2x − 2)
8(x) − 8(1) = 4(2x) − 4(2)
8x − 8 = 8x − 8
− 8x
− 8x
−8 = −8
The equation −8 = −8 is always true. So, the equation has
infinitely many solutions.
49. The principal P is 500. The annual interest rate r is 6%, or
0.06. The interest is compounded yearly. So, n = 1. The
balance y is 800.
r nt
y=P 1+—
n
0.06 1t
800 = 500 1 + —
1
(
)
(
800 =
)
500(1.06)t
Graph y = 800 and y = 500(1.06)t.
1000
47. Because any number to the zero power is 1, the exponent
x − 4 must equal zero. So, x = 4 because 4 − 4 = 0.
48. The initial value is 128. So, a = 128. After each round x, the
1
1
number y of teams gets multiplied by —. So, b = —.
2
2
y = a(b)x
x
()
1
16 = 128( )
2
1
128( )
16
2
=
1
y = 128 —
2
x
Intersection
0 X=8.0661135 Y=800
0
The solution is x ≈ 8.07. So, the balance in the savings
account will be $800 after about 8.07 years.
50. a. The graphs appear to intersect when x = 3. So, the events
will have about the same attendance in 2007.
b. Sample answer: Graph the two equations on a graphing
calculator and use the intersect feature to find the point of
intersection. Another way to check would be to evaluate
each equation for the approximate coordinates (3, 8000)
of the point of intersection. For each equation, the
expressions on each side should be approximately equal
when the coordinates are substituted for the respective
values.
—
x
—
128
—
_______
128
( 12 )
1
1
=( )
2
2
x
1
8
—= —
x
—3
—
2−3 = (2−1)x
2−3 = 2−x
−3 = −x
−x
−3 —
—=
−1 −1
3=x
The solution is x = 3. So, there are 16 teams left after
round 3.
10
51. A base of 1 to any power is 1, even if the powers are not
equal. For example, 13 = 15, but 3 ≠ 5.
52. Yes, it is possible for an exponential equation to have two
different solutions because the graph of the equation formed
with the expression on the left side of the original equation
and the graph of the equation formed with the expression
on the right side of the original equation may intersect
in two points. For example, the equation 2x = x + 3 has
two different solutions x ≈ −2.86 and x ≈ 2.44 because
the graph of y = 2x intersects the graph of y = x + 3 in
two places as shown.
6
−6
6
−2
—
53. 8x−2 = √ 8
8x−2 = 81/2
x−2=
+2
1
2
—
+2
5
x=—
2
(
1
2
1
2
4
2
5
2
—+2=—+—=—
)
5
The solution is x = —.
2
346
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Chapter 6
—
—
57. √ 6
( 61/3 )2x = ( 61/2 )x+6
51/2 = 5x+4
1
—=x+4
2
−4
−4
6(1/3)(2x) = 6(1/2)(x+6)
( 13 )(2x) = ( 12 )(x + 6)
—
(
7
−— = x
2
1
2
1
2
8
2
7
2
—−4=—−—=−—
7
The solution is x = −—.
2
)
1
1
2
2
2
3
1
2
—x = —x + 3
2
3
1
1
− —x − —x
2
2
1
—x = 3
6
1
6 —x = 6 3
6
( 71/5 )x = 72x+3
7(1/5)x = 72x+3
1
— x = 2x + 3
5
⋅
− 2x − 2x
(
1 10
9
9
1
−—x = 3
—−2=—−—=−—
5
5
5
5
5
5
9
5
− — −—x = −— 3
9
5
9
5
x = −—
3
5
The solution is x = −—.
3
⋅( )
56.
⋅
4— x
122x−1 = ( √ 12 )
122x−1
⋅
—−—=—−—=—
4
6
3
6
)
1
2
2
2
1
2
1
2
2
3
1
6
⋅
x = 18
)
The solution is x = 18.
( √5 —3 )5x−10 = ( √8 —3 )4x
58.
( 31/5 )5x−10 = ( 31/8 )4x
3(1/5)(5x−10) = 3(1/8)(4x)
( 15 )(5x − 10) = ( 18 )(4x)
—
1
5
122x−1 = 12(1/4)x
1
2x − 1 = —x
4
− 2x
− 2x
(
( )
(
—
1
2
1
x − 2 = —x
2
−x
−x
1
5
—(5x) − —(10) = —x
= ( 121/4 )x
7
−1 = −—x
4
—
—x = —(x) + —(6)
( √5 —7 )x = 72x+3
55.
—
( 3 )2x = ( √6 )x+6
54. √ 5 = 5x+4
7
1 8
1
— − 2 = — − — = −—
4 4
4
4
4
4
7
−— (−1) = −— −— x
7
7
4
4
—=x
7
4
The solution is x = —.
7
)
1
−2 = −—x
2
−2
⋅ (−2) = −2 ⋅ (
4=x
(
1
2
1
−— x
2
— − 1 = — − — = −—
)
)
The solution is x = 4.
59. yes; If x was 0, b would equal 1. By the definition of negative
x
()
1
exponents, — = (a−1)x. By the Power of a Power Property,
a
(a−1)x = a−x. a must be raised to a positive exponent to stay
positive. So, x must be negative.
Maintaining Mathematical Proficiency
60. Position
Term
1
−20
2
3
4
−26
−32
−38
⤻ ⤻ ⤻
+(−6)
+(−6)
+(−6)
The sequence is arithmetic with a common difference of −6.
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Algebra 1
Worked-Out Solutions
347
Chapter 6
61.
Position
1
2
3
4
Term
9
18
36
72
⤻ ⤻ ⤻
4. Sample answer: If you drop a tennis ball, the height after
each bounce will be a common fraction of the height of the
preceding bounce.
×2
×2
×2
The sequence does not have a common difference. So, it is
not arithmetic.
62.
1. Find the difference between each pair of consecutive terms.
5
Position
1
2
3
4
Term
−5
−8
−12
−17
−3
1
−7
1 − 5 = −4 −3 − 1 = −4 −7 − (−3) = −4
⤻ ⤻ ⤻
+(−3) +(−4) +(−5)
The sequence does not have a common difference. So, it is
not arithmetic.
63.
6.6 Monitoring Progress (pp. 332–335)
The differences are the same. So, the common difference is
−4, and the sequence is arithmetic.
2. Find the ratio between each pair of consecutive terms.
Position
1
2
3
4
Term
10
20
30
40
⤻ ⤻ ⤻
1024
+10
+10
+10
The sequence is arithmetic with a common difference of 10.
6.6 Explorations (p. 331)
1. a.
1
8
128
1024
16
16
128
—=—
1
8
2
2
16
—=—
1
8
—=—
1
The ratios are the same. So, the common ratio is —, and the
8
sequence is geometric.
3. Find the ratio between each pair of consecutive terms.
Step
1
2
3
4
5
Calculator display
2
4
8
16
32
Each number is twice the preceding number.
b.
128
Step
1
2
3
4
5
Calculator display
64
32
16
8
4
2
6
c. Sample answer:
10 5
16 8
6
—=—
—=—
2
6
3
10 5
There is no common ratio. So, the sequence is not geometric.
Find the difference between each pair of consecutive terms.
6
6−2=4
Step
1
2
3
4
5
Calculator display
5
15
45
135
405
1
2
d. The common ratio in part (b) is —. The common ratio in
part (c) is 3.
2. a. When you fold the paper in half once, it will be
0.1(2) = 0.2 millimeter thick. When you fold the paper in
half twice, it will be 0.2(2) = 0.4 millimeter thick. When
you fold the paper in half a third time, it will be
0.4(2) = 0.8 millimeter thick.
b. Sample answer: The most number of times you can fold
a piece of paper in half is 6 times, and the folded paper is
6.4 millimeters thick after 6 folds.
16
—=3
2
Each number is half the preceding number.
10
10
10 − 6 = 4
16
16 − 10 = 6
There is no common difference. So, the sequence is not
arithmetic.
The sequence is neither geometric nor arithmetic.
4. Position 1 2 3
Term
4
5
6
7
1 3 9 27 81 243 729
⤻⤻⤻
×3 ×3 ×3
The next three terms are 81, 243, and 729.
an
800
(7, 729)
600
c. Yes, the statement is true. The thickness is a geometric
sequence that doubles with each fold. After 15 folds, the
thickness is 3276.8 millimeters, or 3.2768 meters, which
is taller than a human being.
3. A geometric sequence describes a pattern when there is a
400
200
(3, 9)
(2, 3)
(1, 1)
2
4
(4, 27)
(6, 243)
(5, 81)
6
8n
common ratio between each consecutive pair of terms. The
sequence is the list of terms that follow the pattern, and the
common ratio describes the sequence.
348
Algebra 1
Worked-Out Solutions
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Chapter 6
5. Position
1
Term
2
2500
3
4
5
6
500 100 20 4
an = a1
0.8 0.16
⤻⤻⤻
1
1
1
×—
×—
rn–1
an = 13(2)n–1
a7 = 13(2)7–1
an = 13(2)n–1
×—
5
5
The next three terms are 4, 0.8, and 0.16.
26
13
9. a1 = 13, r = — = 2
7
= 13(2)6
= 13(64)
5
= 832
The 7th term of the geometric sequence is 832.
an
2
12
(1, 2500)
2400
1
6
10. a1 = 432, r = — = —
3200
an = a1rn–1
(4, 20) (7, 0.16)
(3, 100) (6, 0.8)
1600
800 (2, 500)
()
1
an = 432 —
6
(5, 4)
2
4
—
1
2
3
4
5
Term
80
−40
20
−10
5
6
5
−—
2
7
5
—
4
⤻⤻ ⤻
( ) ( ) ( )
1
1
1
× −— × −— × −—
2
2
2
10
4
11. a1 = 4, r = — = 2.5
an = a1rn–1
The next three terms are 5, −2.5, and 1.25.
)
an =
a7 = 4(2.5)7–1
= 4(2.5)6
4(2.5)n–1
= 4(244.140625)
= 976.5625
(1, 80)
(7, 1.25)
(6, −2.5)
(5, 5)
(3, 20)
2
−40
(4, −10)
The 7th term of the geometric sequence is 976.5625.
12.
8n
1
2560
5
2
3
4
5
6
7
64
−128
⤻⤻ ⤻
×(−2) ×(−2) ×(−2)
The next three terms are −32, 64, and −128.
an
(6, 64)
60
(4, 16)
(2, 4)
2
4
(1, −2)
6
−5
1
an = a1 r n–1
29 = (2)n–1
9=n−1
+1
+1
10 = n
The solution is n = 10. So, the side length of the map is
2560 miles after 10 clicks on the zoom-out button.
Vocabulary and Core Concept Check
1. The first sequence is an arithmetic sequence with a common
difference of 2. The second sequence is a geometric
sequence with a common ratio of 2.
(7, −128)
8. a1 = 1, r = — = −5
512 = (2)n–1
6.6 Exercises (pp. 336–338)
8n
(5, −32)
(3, −8)
−120
5(2)n–1
5
—=—
−2 4 −8 16 −32
Term
y = 5(2)n–1
2560 = 5(2)n–1
(2, −40)
7. Position
−60
6
an = 4(2.5)n–1
an
40
7−1
(
6. Position
80
n–1
1
= 432 —
46,656
1
=—
108
1
The 7th term of the geometric sequence is —.
108
8n
6
n–1
()
()
1
=432( )
6
1
an = 432 —
6
1
a7 = 432 —
6
an = (−5)n–1
a7 = (−5)7–1
an = 1(−5)n–1
= (−5)6
an = (−5)n–1
= 15,625
2. A negative number raised to an odd power is negative, and
raised to an even power is positive. So, if the common ratio
is negative, the terms will alternate signs.
The 7th term of the geometric sequence is 15,625.
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Algebra 1
Worked-Out Solutions
349
Chapter 6
Monitoring Progress and Modeling with Mathematics
9. Find the difference between each pair of consecutive terms.
−8
3.
4
12
36
⤻⤻ ⤻
108
36
36
12
12
4
—=3
—=3
—=3
36
6
⤻⤻⤻
1
6
—=—
36 6
−1
−7
4
10
⤻⤻⤻
1
1
—÷1=—
6
6
1
—
6
16
10. Find the ratio between each pair of consecutive terms.
1
6
—
1
8
0 − (−8) = 8 8 − 0 = 8 16 − 8 = 8
There is a common difference of 8. So, the sequence is
arithmetic.
The common ratio is 3.
4.
0
⤻⤻⤻
108
−7
4
4
−1
1
The common ratio is —.
6
7
4
10
7
10
−7
— = −—
— = −4
— = −—
Find the difference between each pair of consecutive terms.
5.
3
—
8
−3
−192
24
⤻ ⤻ ⤻
−192
24
24
−3
3
−3 ÷ —
8
8
= −3 —
3
= −8
— = −8
⋅
— = −8
0.1
10
100
⤻⤻ ⤻
−7 − 4
10 − (−7)
= −11
= 10 + 7
⋅
= 17
11. Find the ratio between each pair of consecutive terms.
9
100
10
10
1
— = 10
— = 10
14
20
27
⤻⤻⤻
20
14
14
9
10
7
27
20
—=—
—
= 10
—
Find the difference between each pair of consecutive terms.
The common ratio is 10.
128
9
96
72
54
⤻ ⤻⤻
96
128
4 − (−1)
There is neither a common ratio nor a common difference.
So, the sequence is neither geometric nor arithmetic.
1
= 1 10
10
⤻⤻⤻
=5
1 ÷ 0.1
7.
−7
4
=4+1
The common ratio is −8.
6.
−1
3
4
96
128
3
4
54
72
— = —
— = —
3
4
— = —
14
20
27
⤻⤻ ⤻
14 − 9 = 5
20 − 14 = 6
27 − 20 = 7
There is neither a common ratio nor a common difference.
So, the sequence is neither geometric nor arithmetic.
3
The common ratio is —.
4
8.
−162
6
⤻⤻⤻
−18
54
1
The common ratio is −— .
3
1
54
— = −—
3
−162
350
−18
54
1
3
— = −—
Algebra 1
Worked-Out Solutions
6
−18
1
3
— = −—
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Chapter 6
12. Find the ratio between each pair of consecutive terms.
3
7
3
49
—
—
3
16. Find the difference between each pair of consecutive terms.
4
21
⤻⤻⤻
3
3
7 49
3 49
=— —
7 3
=7
—÷—
⋅
3
3÷—
7
7
=3 —
3
=7
11
21
3
4
⋅
11
24
3
8
—
3
⤻⤻⤻
1
3
—=—
24 8
5
20
20 − 5 = 15
120
60 − 20 = 40 120 − 60 = 60
5
20
−4 − (−11)
= −18 + 25
= −11 + 18
= −4 + 11
=7
=7
=7
120
60
—=3
—=2
There is neither a common difference nor a common ratio.
So, the sequence is neither arithmetic nor geometric.
18. Find the difference between each pair of consecutive terms.
−3
6
15
24
⤻⤻ ⤻
6 − (−3)
15 − 6 = 9
24 − 15 = 9
=6+3
=9
15. Find the ratio between each pair of consecutive terms.
There is a common difference of 9. So, the sequence is
arithmetic.
250
250
50
10
—=5
—=5
—=5
10
50
2
There is a common ratio of 5, and the points appear to lie on
an exponential curve. So, the sequence is geometric.
120
60
20
20
5
—=4
There is a common difference of 7. So, the sequence is
arithmetic.
⤻⤻⤻
60
⤻ ⤻⤻
−4
−18 − (−25) −11 − (−18)
50
60
Find the ratio between each pair of consecutive terms.
⤻⤻⤻
10
16
⤻ ⤻⤻
3
— ÷3
8
3 1
=— —
8 3
1
=—
8
−11
—
11
17. Find the difference between each pair of consecutive terms.
14. Find the difference between each pair of consecutive terms.
2
—
There is neither a common difference nor a common ratio.
So, the sequence is neither arithmetic nor geometric.
1
There is a common ratio of — . So, the sequence is
8
geometric.
−18
19
⤻⤻
⤻
19
16
11
4
⋅
−25
19 − 16 = 3
16
—
13. Find the ratio between each pair of consecutive terms.
1
24
—=—
192 8
16 − 11 = 5
Find the ratio between each pair of consecutive terms.
There is a common ratio of 7. So, the sequence is geometric.
192
19
⤻ ⤻⤻
11 − 4 = 7
—=7
16
19.
Position
1
2
3
4
5
6
7
Term
5
20
80
320
1280
5120
20,480
⤻
⤻
⤻
×4
×4
×4
The next three terms are 1280, 5120, and 20,480.
an
24,000
(7, 20,480)
18,000
12,000
6000
(4, 320)
(3, 80)
(2, 20)
2
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(6, 5120)
(5, 1280)
(1, 5)
4
6
8n
Algebra 1
Worked-Out Solutions
351
Chapter 6
20. Position
Term
1
2
3
−3
12
4
5
6
23. Position
7
−48 192 −768 3072 −12,288
Term
⤻⤻⤻
1
2
3
32
8
2
4
1
—
2
(3, −48)
(2, 12)
(4, 10)
(6, 3072)
4
−5000 (1, −3)
(1, 32)
8n
6
24
(5, −768)
16
−10,000
8
(7, −12,288)
−15,000
(2, 8)
(5, 18) (7, 1128)
(4, 12) (6, 312)
(3, 2)
2
Position
1
2
3
4
5
6
7
Term
81
−27
9
−3
1
1
−—
3
—
1
9
4
24. Position
8n
6
1
2
3
4
5
6
7
16
9
—
8
3
4
6
9
—
27
2
—
3
×—
2
×—
2
×—
2
—
Term
1
1
The next three terms are 1, −—, and — .
3
9
(1, 81)
(7,
(5, 1)
(4, −3)
1
(6, − 3)
(3, 9)
2
6
)
1
9
27
81
The next three terms are 9, —, and — .
2
4
an
8n
24
(2, −27)
−40
(7, 841)
18
22. Position
1
2
−375 −75
Term
3
−15
4
5
−3
6
12
7
3
3
−— −—
25
125
1
1
×— ×—
5
5
6
3
−—
5
6
−300
−400
4
6
8
2
25. a1 = 2, r = — = 4
an =
an =
a1rn−1
2(4)n−1
(7, −1235)
(3, −15)
(6, −235)
(4, −3)
(5, −35)
(1, −375)
an = 2(4)n−1
a6 = 2(4)6−1
= 2(4)5
= 2048
The 6th term of the geometric sequence is 2048.
26.
a1 = 0.6,
15
r = — = −5
−3
an = a1rn−1
an = 0.6(−5)n−1
an
8n
= 2(1024)
8n
−100 (2, −75)
−200
(5, 9)
(4, 6)
(3, 4)
2
3
3
3
The next three terms are −— , −— , and −— .
125
5
25
4
(6, 227)
(1, 196 )
(2, 83 )
⤻
⤻⤻
1
×—
5
2
81
4
⤻⤻
⤻
3
3
an
40
×—
4
an
32
2
80
7
1
—
128
1
1 1
The next three terms are — , — , and — .
8 32
128
10,000
21.
×—
4
×—
4
The next three terms are −768, 3072, and −12,288.
an
6
1
—
32
⤻
⤻
⤻
1
1
1
×(−4) ×(−4) ×(−4)
5000
5
1
—
8
an = 0.6(−5)n−1
a6 = 0.6(−5)6−1
= 0.6(−5)5
= 0.6(−3125)
= −1875
The 6th term of the geometric sequence is −1875.
352
Algebra 1
Worked-Out Solutions
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Chapter 6
1
8
1
= −—
4
1
4
( 18 )
1
an = −— (2)n−1
8
1 6−1
a6 = −— (2)
8
1
= −— (2)5
8
1
= −— (32)
8
27. a1 = −—, r = −— ÷ −—
⋅ (−8)
=2
an = a1rn−1
1
an = −— (2)n−1
8
32. a1 = 224
28 1
r=—=—
56 2
0.9
r=—=9
0.1
an = a1rn−1
an = 0.1(9)n−1
an = 0.1(9)n−1
a6 = 0.1(9)6−1
33.
Round
Number of teams
an = 7640(0.1)n−1
= 7640(0.000001)
= 0.0764
n−1
( )
( )
( )
1
= −192( −
1024 )
1
an = −192 −—
4
1
a6 = −192 −—
4
1
= −192 −—
4
6−1
5
3
=—
16
3
The 6th term of the geometric sequence is —.
16
an = a1 rn−1
18
r = — = −6 an = 0.5(−6)n−1
−3
an = 0.5(−6)n−1
a6 = 0.5(−6)6−1
= 0.5(−6)5
= 0.5(−7776)
= −3888
The 6th term of the geometric sequence is −3888.
2
32
3
16
4
8
5
4
⤻
⤻
⤻
⤻
⤻
1
1
1
1
1
34. Zoom outs
Display area
(square units)
×—
2
×—
2
0
1
2
3
4
96
384
1536
6144
24,576
⤻⤻⤻⤻
×4
n−1
—
31. a1 = 0.5
1
64
×— ×—
2
2
1
The geometric sequence has a common ratio of —, and the
2
next three terms are 16, 8, and 4. So, after the third, fourth,
and fifth rounds the number of teams that remain is 16, 8,
and 4, respectively.
The 6th term of the geometric sequence is 0.0764.
( )
0
128
×—
2
764
r = — = 0.1 an = 7640(0.1)n−1 a6 = 7640(0.1)6−1
7640
= 7640(0.1)5
1
3
1
r = — = −— an = −192 −—
−12
4
4
5
=7
The 6th term of the geometric sequence is 5904.9.
an = a1rn−1
6−1
The 6th term of the geometric sequence is 7.
= 0.1(9)5
an = a1rn−1
—
6
n−1
—
= 5904.9
30. a1 = −192
n−1
= −4
= 0.1(59049)
29. a1 = 7640
()
1
an = 224 —
2
—
The 6th term of the geometric sequence is −4.
28. a1 = 0.1
()
1
a = 224( )
2
1
= 224( )
2
1
= 224( )
32
1
an = 224 —
2
an = a1 rn−1
×4
×4
×4
1
The geometric sequence has a common ratio of —, and the
2
next two terms are 6144 and 24,576. So, after you zoom
out four times, the screen displays an area of 24,576 square
units.
1
2
35. The common factor is −—, not −2.
−8,
4,
−2,
1, . . .
⤻ ⤻ ⤻
( ) ( ) ( )
1
1
1
× −— × −— × −—
2
2
2
1
1 1
The next three terms are −—, —, and −—.
2 4
8
36. The common ratio is 6, not −6.
−2, −12, −72, −432
⤻ ⤻ ⤻
×6
×6
×6
The first term is −2, and the common ratio is 6.
an = a1 rn−1
an = −2(6)n−1
So, an equation for the nth term of the geometric sequence is
an = −2(6)n−1.
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Algebra 1
Worked-Out Solutions
353
Chapter 6
37. a. Swing
Distance (in millimeters)
1
625
2
500
3
400
39. a. The first figure has 1 square, and the number of small
squares contained in the big square has a common ratio of
9 for each pair of consecutive figures.
⤻
⤻
4
4
×—
5
f(n) = a1 rn−1
×—
5
f(n) = 1(9)n−1
4
The first term is 625, and the common ratio is —.
5
f(n) = a1 rn−1
4 n−1
f(n) = 625 —
5
f(n) = 9n−1
So, a function that represents how many small squares
are contained in the large square of the nth figure of the
sequence is f(n) = 9n−1.
()
A function that represents the distance the pendulum
4 n−1
.
swings on its nth swing is f(n) = 625 —
5
4 n−1
b. f(n) = 625 —
5
()
()
4
256 = 625( )
5
4
625( )
256
5
=
—
n−1
= 99
= 387,420,489
So, the 10th figure will be a large square containing
387,420,489 small squares.
40. a. The first figure has 1 triangle, and the number of small
triangles contained in the big triangle has a common ratio
of 4 for each pair of consecutive figures.
625
625
4 n−1
256
—= —
5
625
4 n−1
44
—= —
5
54
()
()
( 45 ) = ( 45 )
—
4
—
+1
5=n
The solution is n = 5. So, after 5 swings, the distance is
256 millimeters.
38. a. The first term is 6, and the common ratio is 6.
f(n) = a1 rn−1
f(n) = 6(6)n−1
f(n) =
61
⋅
6n−1
f(n) = 61+n−1
f(n) = 6n
A function that represents the number of people who
receive the email on the nth day is an = 6n.
b.
f(n) = a1 rn−1
f(n) = 1(4)n−1
f(n) = 4n−1
So, a function that represents how many small triangles
are contained in the large triangle of the nth figure of the
sequence is f(n) = 4n−1.
n−1
4=n−1
+1
f(n) = 6n
1296 = 6n
64 = 6n
4=n
f(n) = 9n−1
f(10) = 910−1
n−1
—
_________
—
b.
b.
f(n) = 4n−1
f(10) = 410−1
= 49
= 262,144
So, the 10th figure will be a large triangle containing
262,144 small triangles.
41. Round, n
1
2
Number
128−64 64−32
of teams
= 32
eliminated = 64
3
4
5
32−16
= 16
16−8
=8
8−4
=4
⤻ ⤻ ⤻ ⤻
1
×—
2
1
×—
2
1
×—
2
1
×—
2
1
The sequence is geometric since it has a common ratio of —.
2
So, a sequence that represents the number of teams that have
()
1
been eliminated after n rounds is an = 64 —
2
n−1
.
The solution is n = 4. So, on the 4th day, 1296 people will
have received the email.
354
Algebra 1
Worked-Out Solutions
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Chapter 6
42. From the figure in Exercise 34, you can see that the original
screen display is 12 units long and 8 units wide. If you
repeatedly multiply each of these dimensions by 2, you
get the areas that match the sequence described. So, the
sequence of perimeters are as follows in the table.
0.02
0.01
50. a. The first term is 0.01, and the common ratio is — = 2.
an = a1r n−1
an = 0.01(2)n−1
An equation that represents the nth term of the geometric
sequence is an = 0.01(2)n−1.
Zoom outs
0
1
2
Length (units)
12
24
48
Width (units)
8
16
32
a25 = 0.01(2)25−1
Display area (square units)
96
384
1536
= 0.01(2)24
Perimeter (units)
40
80
160
= 0.01(16,777,216)
80 160
Because — = — = 2, the sequence has a common ratio
40
80
of 2. So, it is geometric. A sequence that represents the
perimeter of the graphing calculator screen after you zoom
out n times is an = 80(2)n−1.
43. Sample answer: Graphs of arithmetic sequences form
a linear pattern. Graphs of geometric sequences form a
curve. When the common ratio is positive, the graph forms
an exponential curve pattern. When the common ratio is
negative, the graph forms a pattern of points alternating
between quadrants I and IV (or II and III).
44. Your friend is correct. This cannot be a geometric sequence.
The only way to get 0 as a term would be with a common
ratio of 0, but if the common ratio is 0, then there is no
possible previous term that would result in −8 as the
next term.
45. The sequence shown is both an arithmetic and a geometric
sequence. It is an arithmetic sequence with a common
difference of 0 and a geometric sequence with a common
ratio of 1.
46. a. B; When the common ratio is greater than 1, the values in
the sequence increase.
b. A; When the common ratio is between 0 and 1, the values
in the sequence decrease.
47. an = a1r n−1
81 = a1(3)3−1
an = a1r n−1
an = 9(3)n−1
81 = a1(3)2
81 = a1(9)
a9 = 9(3)9−1
a (9)
9
= 9(3)8
= 9(6561)
= 59,049
81
9
1
—=—
9 = a1
The 9th term of the geometric sequence is 59,049.
b. an = 0.01(2)n−1
= 167,772.16
So, she will pay $167,772.16 on the 25th day of
the month.
c. The college student should have chosen to live on campus
because she is paying millions of dollars by the end of the
month with this plan.
16 gal
1
51. a. —
4 qt 2 pt 2 c 8 fl oz 16(4)(2)(2)(8) fl oz
— — — = ——
⋅—
1 gal ⋅ 1 qt ⋅ 1 pt ⋅ 1 c
1
= 2048 fl oz
1
1 3
If — of the soup is served, then 1 − — = — of the soup is
4
4 4
left after each day.
3
After the first day, 2048 — = 1536 fluid ounces of soup
4
are left.
()
1536,
1152,
864,
648,
486
⤻3
⤻
⤻
⤻
3
3
3
×—
4
×—
4
×—
4
×—
4
The first five terms of the sequence of the number of fluid
ounces of soup left each day are 1536, 1152, 864, 648,
and 486.
3
b. The first term is 1536, and the common ratio is —.
4
an = a1r n−1
()
3
an = 1536 —
4
n−1
So, an equation that represents the nth term of the
3 n−1
.
sequence is an = 1536 —
4
()
3
4
of the previous day’s amount will still remain. So, the
amount of soup will get very small, but it will never run
out. In other words, an exponential function approaches
0, but never actually equals 0. However, in real life,
eventually there will not be enough soup to serve.
c. Theoretically, the soup will never run out. Each day, —
48. Sample answer: 1, 1, 2, 3, 5, 8, 13, . . .; After the first
two terms, each term is the sum of the preceding two terms.
49. The terms of a geometric sequence are dependent because
each term is calculated from the preceding term.
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Algebra 1
Worked-Out Solutions
355
Chapter 6
52. Putting the numbers in a spreadsheet and calculating the sum
after each term is added shows that the sum approaches 2.
54. In the first table, a1 = 2, and the common ratio is 3.
an = a1r n−1
1
an = 2(3)n−1
0.5
In the second table, b1 = 1, and the common ratio is 5.
0.25
bn = b1r n−1
0.125
bn = 1(5)n−1
0.0625
bn = 5n−1
0.03125
0.007813
Equations that represent the nth term of each geometric
sequence shown are an = 2(3)n−1 and bn = 5n−1.
0.003906
a. a1 − b1 = 2 − 1
0.015625
=1
0.000977
a3 − b3 = 18 − 25
0.000488
a4 − b4 = 54 − 125
= −71
The differences form the sequence 1, 1, −7, −71, . . ..
This sequence does not have a common ratio. So, it is not
a geometric sequence.
0.000122
6.1E-05
1.999939
a
b2
a
18
54
—4 = —
25
b4 125
6 18 54
The ratios form the sequence 2, —, —, —, . . .
5 25 125
Find the ratio between each pair of consecutive terms.
2
1
a
b1
b. —1 = — = 2
Sample answer: Another infinite geometric sequence that has
the same sum is
1 3 9 27
2 8 32 128
=1
= −7
0.000244
()
1 3
2 4
—, —, —, —, …, — —
n−1
,…
6
5
6
=—
5
3
=—
5
0.375
0.28125
0.210938
0.158203
0.088989
0.066742
0.037542
0.028157
0.021118
0.015838
0.011879
0.008909
1.973273
53. Sample answer: 1, −2, 4, −8, …
Algebra 1
Worked-Out Solutions
18
25
18
=—
25
3
=—
5
—3 = —
6
5
5
—
6
54
125
54
=—
125
3
=—
5
—÷—
⋅ —12
⋅
18
25
25
—
18
—÷—
⋅
The ratios are the same. So, the sequence is geometric.
3
The common ratio — of this sequence is the quotient of the
5
common ratios of the two sequences from the tables.
0.118652
0.050056
a
b3
6
5
—2 = —
—÷2
0.5
356
a2 − b2 = 6 − 5
0.001953
Maintaining Mathematical Proficiency
55.
x
y
y-Value from
model
Residual
0
−10
−8
−10 − (−8) = −2
1
−2
−5
−2 − (−5) = 3
2
−1
−2
−1 − (−2) = 1
3
2
1
2−1=1
4
1
4
1 − 4 = −3
5
7
7
7−7=0
6
10
10
10 − 10 = 0
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Chapter 6
2.
residual
4
a1
1
a2
1
a3
2
a4
3
a5
5
a6
8
a7
13
a8
21
2
This sequence is the same as that of Exploration 1.
2
4
x
6
3. Give the beginning term(s) of a sequence and a recursive
−2
equation that tells how an is related to one or more
preceding terms.
−4
The residual points are evenly dispersed about the horizontal
axis. So, the equation y = 3x − 8 is a good fit.
56.
4. The sequence described in Explorations 1 and 2 are the
Fibonacci numbers, named after Leonardo Fibonacci, a
famous Italian mathematician who lived during the
Middle Ages. He was the first mathematician to publish a
description of these numbers.
x
y
y-Value from
model
Residual
−3
6
16
6 − 16 = −10
−2
4
11
4 − 11 = −7
−1
6
6
6−6=0
0
1
1
1−1=0
1
2
−4
2 − (−4) = 6
a5 = a4 − 8 = −24 − 8 = −32
2
−4
−9
−4 − (−9) = 5
a6 = a5 − 8 = −32 − 8 = −40
3
−3
−14
−3 − (−14) = 11
4
6.7 Monitoring Progress (pp. 341−343)
1. a1 = 0
a2 = a1 − 8 = 0 − 8 = −8
a3 = a2 − 8 = −8 − 8 = −16
a4 = a3 − 8 = −16 − 8 = −24
The first six terms of the sequence are 0, −8, −16, −24,
−32, and −40.
2
residual
4
8n
6
−10
2
−20
−4
−2
2
4 x
−30
−2
−40
an
−4
The residual points are not evenly dispersed about the
horizontal axis. So, the equation y = −5x + 1 is not a
good fit.
2. a1 = −7.5
a2 = a1 + 2.5 = −7.5 + 2.5 = −5
a3 = a2 + 2.5 = −5 + 2.5 = −2.5
a4 = a3 + 2.5 = −2.5 + 2.5 = 0
6.7 Explorations (p. 339)
1. Months
Number
of pairs
1
1
2
1
3
2
4
3
5
5
6
8
7
13
8
21
a5 = a4 + 2.5 = 0 + 2.5 = 2.5
a6 = a5 + 2.5 = 2.5 + 2.5 = 5
The first six terms of the sequence are −7.5, −5, −2.5, 0,
2.5, and 5.
8
an
4
2
4
6
8n
−4
In month 6, there are 8 breeding pairs, in month 7, there are
13 breeding pairs, and in month 8, there are 21 breeding pairs.
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−8
Algebra 1
Worked-Out Solutions
357
Chapter 6
3. a1 = −36
6. Position, n
1
1
a2 = — a1 = — (−36) = −18
2
2
1
1
a3 = — a2 = — (−18) = −9
2
2
9
1
1
a4 = — a3 = — (−9) = −—
2
2
2
1
1 9
9
a5 = — a4 = — −— = −—
2
2 2
4
1
1 9
9
a6 = — a5 = — −— = −—
2
2 4
8
Term, an
4
5
2.6
3.9
5.2
6.5
⤻ ⤻ ⤻ ⤻
+1.3
+1.3
+1.3
an = an−1 + d
an = an−1 + 1.3
So, a recursive rule for the sequence is
8n
6
3
The sequence is arithmetic, with first term a1 = 1.3 and
common difference d = 1.3.
9
The first six terms of the sequence are −36, −18, −9, −—,
2
9
9
−—, and −—.
4
8
4
1.3
2
+1.3
( )
( )
2
1
a1 = 1.3, an = an−1 + 1.3.
7. Position, n
Term, an
1
2
3
4
5
4
20
100
500
2500
⤻ ⤻ ⤻ ⤻
×5
−10
×5
×5
×5
The sequence is geometric, with first term a1 = 4 and
common ratio r = 5.
−20
⋅
an = r an−1
−30
an = 5an−1
−40
an
So, a recursive rule for the sequence is a1 = 4, an = 5an−1.
4. a1 = 0.7
8. Position, n
a2 = 10a1 = 10(0.7) = 7
Term, an
a3 = 10a2 = 10(7) = 70
1
128
2
3
4
5
−32
8
−2
0.5
⤻ ⤻ ⤻ ⤻
( ) ×( −14 ) ×( −14 ) ×( −14 )
1
× −—
4
a4 = 10a3 = 10(70) = 700
a5 = 10a4 = 10(700) = 7000
a6 = 10a5 = 10(7000) = 70,000
The first six terms of the sequence are 0.7, 7, 70, 700, 7000,
and 70,000.
an
—
—
—
The sequence is geometric, with first term a1 = 128 and
1
common ratio r = −—.
4
⋅
an = r an−1
1
an = −— an−1
4
80,000
So, a recursive rule for the sequence is
1
a1 = 128, an = −— an−1.
4
60,000
40,000
20,000
9. Month, n
2
5. Position, n
Term, an
4
6
1
8
Height (feet), an
8n
2
3
4
5
3
−2
−7
−12
⤻ ⤻ ⤻ ⤻
+(−5) +(−5) +(−5) +(−5)
The sequence is arithmetic, with first term a1 = 8 and
common difference d = −5.
an = an−1 + d
an = an−1 + (−5)
1
2
3
4
2
3.5
5
6.5
⤻ ⤻ ⤻
+1.5
+1.5
+1.5
The sequence is arithmetic, with first term a1 = 2 and
common difference d = 1.5.
an = an−1 + d
an = an−1 + 1.5
So, a recursive rule for the sequence is
a1 = 2 feet, an = (an−1 + 1.5) feet.
an = an−1 − 5
So, a recursive rule for the sequence is
a1 = 8, an = an−1 − 5.
358
Algebra 1
Worked-Out Solutions
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Chapter 6
10. The recursive rule represents an arithmetic sequence, with
first term a1 = −45 and common difference d = 20.
Use the recursive equation an = an−2 + an−1 to find the next
three terms.
an = a1 + (n − 1)d
a6 = a4 + a5 = 17 + 28 = 45
an = −45 + (n−1)(20)
a7 = a5 + a6 = 28 + 45 = 73
an = −45 + n(20) − 1(20)
a8 = a6 + a7 = 45 + 73 = 118
an = −45 + 20n − 20
The next three terms are 45, 73, and 118.
an = 20n − 65
15. Find the sum of each pair of consecutive terms.
An explicit rule for the sequence is an = 20n − 65.
a1 + a2 = −3 + −4 = −7
11. The recursive rule represents a geometric sequence, with first
term a1 = 13 and common ratio r = −3.
a3 + a4 = −7 + −11 = −18
an = a1r n−1
Beginning with the third term, each term is the sum of the
two previous terms. So, a recursive rule for the sequence is
a1 = −3, a2 = −4, an = an−2 + an−1.
an = 13(−3)n−1
An explicit rule for the sequence is an = 13(−3)n−1.
12. The explicit rule represents an arithmetic sequence, with first
term a1 = −(1) + 1 = 0 and common difference d = −1.
an = an−1 + d
The next three terms are −29, −47, and −76.
So, a recursive rule for the sequence is
a1 = 0, an = an−1 − 1.
16. Find the difference of each term and the one before it.
13. The explicit rule represents a geometric sequence, with first
term a1 = −2.5 and common difference r = 4.
a6 − a5 = 0 − (−1) = 1
a7 − a6 = 1 − 0 = 1
a1 = −2.5, an = 4an−1.
6
6−5=1
5
11
11 − 6 = 5
6
6
5
—
a3 − a2 = 0 − 1 = −1
a5 − a4 = −1 − (−1) = 0
So, a recursive rule for the sequence is
5
a2 − a1 = 1−1 = 0
a4 − a3 = −1 − 0 = −1
n−1
an = 4an−1
14.
a6 = a4 + a5 = −11 + (−18) = −29
a8 = a6 + a7 = −29 + (−47) = −76
an = an−1 − 1
⋅a
Use the recursive equation an = an−2 + an−1 to find the next
three terms.
a7 = a5 + a6 = −18 + (−29) = −47
an = an−1 + (−1)
an = r
a2 + a3 = −4 + −7 = −11
17
28
17 − 11 = 6 28 − 17 = 11
11
11
6
—
17
17
11
—
28
28
17
—
The sequence has neither a common difference nor a
common ratio. So, the sequence is neither arithmetic nor
geometric. Then, find the sum of each pair of consecutive
terms.
Beginning with the third term, each term is the difference of
the previous term and the one before that. So, a recursive rule
for the sequence is a1 = 1, a2 = 1, an = an−1 − an−2.
Use the recursive equation an = an−1 − an−2 to find the next
three terms.
a9 = a8 − a7 = 1−1 = 0
a10 = a9 − a8 = 0 − 1 = −1
a11 = a10 − a9 = −1 − 0 = −1
The next three terms are 0, −1, and −1.
a1 + a2 = 5 + 6 = 11
a2 + a3 = 6 + 11 = 17
a3 + a4 = 11 + 17 = 28
Beginning with the third term, each term is the sum of the
two previous terms. So, a recursive rule for the sequence is
a1 = 5, a2 = 6, an = an−2 + an−1.
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Algebra 1
Worked-Out Solutions
359
Chapter 6
17. Find the difference of each pair of consecutive terms.
a1 − a2 = 4 − 3 = 1
8. a1 = 10
a2 = a1 − 5 = 10 − 5 = 5
a2 − a3 = 3 − 1 = 2
a3 = a2 − 5 = 5 − 5 = 0
a3 − a4 = 1 − 2 = −1
a4 = a3 − 5 = 0 − 5 = −5
a4 − a5 = 2 − (−1) = 3
a5 = a4 − 5 = −5 − 5 = −10
a5 − a6 = −1 − 3 = −4
a6 = a5 − 5 = −10 − 5 = −15
Beginning with the third term, each term is the difference of
the two previous terms. So, a recursive rule for the sequence
is a1 = 4, a2 = 3, an = an−2 − an−1.
The first six terms of the sequence are 10, 5, 0, −5, −10,
and −15.
Use the recursive equation an = an−2 − an−1 to find the next
three terms.
16
an
8
a8 = a6 − a7 = 3 − (−4) = 7
a9 = a7 − a8 = −4 − 7 = −11
2
a10 = a8 − a9 = 7 − (−11) = 18
−16
6.7 Exercises (pp. 344–346)
9. a1 = 2
Vocabulary and Core Concept Check
a2 = 3a1 = 3(2) = 6
1. A recursive rule gives the beginning term(s) of a sequence
and a recursive equation that tells how an is related to one or
more preceding terms.
2. The rule that does not belong is an = 6n − 2 because it is the
only explicit rule. The other three are recursive.
Monitoring Progress and Modeling with Mathematics
represents a geometric sequence.
⋅a
n−1·
So, it
represents an arithmetic sequence.
a4 = 3a3 = 3(18) = 54
a5 = 3a4 = 3(54) = 162
a6 = 3a5 = 3(162) = 486
The first six terms of the sequence are 2, 6, 18, 54, 162,
and 486.
an
450
300
5. The rule an = an−1 − 4 is of the form an = an−1 + d. So, it
represents an arithmetic sequence.
6. The rule an = −6an−1 is of the form an = r
a3 = 3a2 = 3(6) = 18
600
4. The rule an = an−1 + 1 is of the form an = an−1 + d. So, it
represents a geometric sequence.
8n
6
−8
The next three terms are 7, −11, and 18.
3. The rule an = 7an−1 is of the form an = r
4
⋅a
n−1·
So, it
150
2
4
6
8n
10. a1 = 8
a2 = 1.5a1 = 1.5(8) = 12
7. a1 = 0
a2 = a1 + 2 = 0 + 2 = 2
a3 = a2 + 2 = 2 + 2 = 4
a4 = a3 + 2 = 4 + 2 = 6
a5 = a4 + 2 = 6 + 2 = 8
a6 = a5 + 2 = 8 + 2 = 10
The first six terms of the sequence are 0, 2, 4, 6, 8, and 10.
an
a3 = 1.5a2 = 1.5(12) = 18
a4 = 1.5a3 = 1.5(18) = 27
a5 = 1.5a4 = 1.5(27) = 40.5
a6 = 1.5a5 = 1.5(40.5) = 60.75
The first six terms of the sequence are 8, 12, 18, 27, 40.5,
and 60.75.
an
80
16
60
12
40
8
20
4
2
2
360
4
6
4
6
8n
8n
Algebra 1
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Chapter 6
14.
11. a1 = 80
1
1
a2 = −— a1 = − — (80) = −40
2
2
1
1
a3 = −— a2 = − — (−40) = 20
2
2
1
1
a4 = −— a3 = − — (20) = −10
2
2
1
1
a5 = −— a4 = − — (−10) = 5
2
2
5
1
1
a6 = −— a5 = − — (5) = − —
2
2
2
The first six terms of the sequence are 80, −40, 20, −10, 5,
5
and − —.
2
1
2
3
4
Term, an
8
24
72
216
⤻
⤻
⤻
×3
×3
×3
The sequence is geometric, with first term a1 = 8 and
common ratio r = 3.
an = r
⋅a
n−1
an = 3an−1
So, a recursive rule for the sequence is a1 = 8, an = 3an−1·
15.
Position, n
Term, an
an
1
2
3
4
5
243
81
27
9
3
⤻
⤻
⤻
⤻
1
1
1
1
×—
×—
×—
×—
3
3
3
3
The sequence is geometric, with first term a1 = 243 and
1
common ratio r = —.
3
an = r an−1
1
an = — an−1
3
1
So, a recursive rule for the sequence is a1 = 243, an = — an−1·
3
80
40
⋅
8n
2
Position, n
−40
12. a1 = −7
a2 = −4a1 = −4(−7) = 28
16.
a3 = −4a2 = −4(28) = −112
a4 = −4a3 = −4(−112) = 448
a5 = −4a4 = −4(448) = −1792
Position, n
1
2
3
4
5
Term, an
3
11
19
27
35
⤻
⤻
⤻
⤻
+8
+8
+8
+8
The sequence is arithmetic, with first term a1 = 3 and
common difference d = 8.
a6 = −4a5 = −4(−1792) = 7168
The first six terms of the sequence are −7, 28, −112, 448,
−1792, and 7168.
an = an−1 + d
an = an−1 + 8
an
So, a recursive rule for the sequence is a1 = 3, an = an−1 + 8.
6000
17.
3000
2
4
6
1
Term, an
0
8n
−3000
13.
Position, n
2
3
4
5
−3
−6
−9
−12
⤻ ⤻ ⤻ ⤻
+(−3) +(−3) +(−3) +(−3)
The sequence is arithmetic, with first term a1 = 0 and
common difference d = −3.
Position, n
1
2
3
4
an = an−1 + d
Term, an
7
16
25
34
an = an−1 + (−3)
⤻
⤻
⤻
+9
+9
+9
The sequence is arithmetic, with first term a1 = 7 and
common difference d = 9.
an = an−1 + d
an = an−1 + 9
an = an−1 − 3
So, a recursive rule for the sequence is a1 = 0, an = an−1 − 3.
18.
Position, n
1
Term, an
5
So, a recursive rule for the sequence is a1 = 7, an = an−1 + 9.
2
3
4
5
−20
80
−320
1280
⤻ ⤻ ⤻ ⤻
×(−4) ×(−4) ×(−4) ×(−4)
The sequence is geometric, with first term a1 = 5 and
common ratio r = −4.
an = r
⋅a
n−1
an = −4an−1
So, a recursive rule for the sequence is a1 = 5, an = −4an−1·
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Algebra 1
Worked-Out Solutions
361
Chapter 6
19.
Position, n
Term, an
1
2
3
4
−1
−4
−16
−64
23. The recursive rule represents an arithmetic sequence, with
first term a1 = −3 and common difference d = 3.
an = a1 + (n − 1)d
⤻
⤻
⤻
×4
×4
×4
an = −3 + (n − 1)3
The sequence is geometric, with first term a1 = −1 and
common ratio r = 4.
an = r
an = −3 + n(3) − 1(3)
an = −3 + 3n − 3
⋅a
an = 3n − 6
n−1
an = 4an−1
So, a recursive rule for the sequence is a1 = −1, an = 4an−1·
20.
Position, n
1
2
3
4
Term, an
35
24
13
2
⤻ ⤻ ⤻
an = a1 + (n − 1)d
an = 8 + (n − 1)(−12)
an = 8 + n (−12) − 1(−12)
an = 8 − 12n + 12
an = −12n + 20
an = an−1 + d
An explicit rule for the sequence is an = −12n + 20.
an = an−1 + (−11)
an = an−1 −11
25. The recursive rule represents a geometric sequence, with first
So, a recursive rule for the sequence is a1 = 35, an = an−1 − 11.
term a1 = 16 and common ratio r = 0.5.
an = a1r n−1
Hour, n
1
2
3
4
an = 16(0.5)n−1
Bacterial cells, an
1
2
4
8
An explicit rule for the sequence is an = 16(0.5)n−1.
⤻
⤻
⤻
×2
×2
×2
26. The recursive rule represents a geometric sequence, with first
The sequence is geometric, with first term a1 = 1 and
common ratio r = 2.
term a1 = −2 and common ratio r = 9.
an = r
an = −2(9)n−1
an = a1r n−1
⋅a
n−1
an = 2an−1
An explicit rule for the sequence is an = −2(9) n−1.
So, a recursive rule for the sequence is
27. The recursive rule represents an arithmetic sequence, with
a1 = 1 cell, an = 2an−1 cells.
22.
24. The recursive rule represents an arithmetic sequence, with
first term a1 = 8 and common difference d = −12.
+(−11) +(−11) +(−11)
The sequence is arithmetic, with first term a1 = 35 and
common difference d = −11.
21.
An explicit rule for the sequence is an = 3n − 6.
Day, n
1
1
4—
2
Length (inches), an
first term a1 = 4 and common difference d = 17.
2
3
4—
4
3
5
⤻ ⤻ ⤻
1
1
1
+—
+—
+—
4
4
4
1
The sequence is arithmetic, with first term a1 = 4 — and
2
1
common difference d = —.
4
an = an−1 + d
1
an = an−1 + —
4
So, a recursive rule for the sequence is
(
)
1
1
a1 = 4 — inches, an = an−1 + — inches.
4
2
4
1
5—
4
an = a1 + (n − 1)d
an = 4 + (n − 1)17
an = 4 + n(17) − 1(17)
an = 4 + 17n − 17
an = 17n − 13
An explicit rule for the sequence is an = 17n − 13.
28. The recursive rule represents a geometric sequence, with first
term an = 5 and common ratio r = −5.
an = a1r n−1
an = 5(−5)n−1
An explicit rule for the sequence is an = 5(−5) n−1.
29. The explicit rule represents a geometric sequence, with first
term a1 = 7 and common ratio r = 3.
an = r
⋅a
n−1
an = 3an−1
So, a recursive rule for the sequence is a1 = 7, an = 3an−1·
362
Algebra 1
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Chapter 6
30. The explicit rule represents an arithmetic sequence, with first
36.
term a1 = −4(1) + 2 = −2 and common difference d = −4.
an
16
an = an−1 + d
12
an = an−1 + (−4)
an = an−1 − 4
So, a recursive rule for the sequence is a1 = −2, an = an−1 − 4.
8
4
31. The explicit rule represents an arithmetic sequence, with first
term a1 = 1.5(1) + 3 = 4.5 and common difference d = 1.5.
an = an−1 + d
The sequence is geometric, with first term a1 = 16 and
1
common ratio r = —.
2
an = r an−1
1
an = — an−1
2
1
So, a recursive rule for the sequence is a1 = 16, an = — an−1·
2
an = a1r n−1
1 n−1
an = 16 —
2
1 n−1
An explicit rule for the sequence is an = 16 —
.
2
an = an−1 + 1.5
⋅
So, a recursive rule for the sequence is
a1 = 4.5, an = an−1 + 1.5.
32. The explicit rule represents an arithmetic sequence, with first
term a1 = 6(1) − 20 = −14 and common difference d = 6.
an = an−1 + d
()
an = an−1 + 6
So, a recursive rule for the sequence is
a1 = −14, an = an−1 + 6.
33. The explicit rule represents a geometric sequence, with first
37.
30
term a1 = 1 and common ratio r = −5.
20
an = r
10
⋅a
n−1
4n
2
an
an = −5an−1
So, a recursive rule for the sequence is a1 = 1, an = −5an−1·
()
2
4n
−10
34. The explicit rule represents a geometric sequence, with first
2
term a1 = −81 and common ratio r = —.
3
an = r an−1
2
an = — an−1
3
2
So, a recursive rule for the sequence is a1 = −81, an = — an−1·
3
⋅
35.
The sequence is geometric, with first term a1 = −1 and
common ratio r = −3.
an = r
⋅a
n−1
an = −3an−1
So, a recursive rule for the sequence is a1 = −1, an = −3an−1·
an
an = a1r n−1
80
an = −1(−3)n−1
60
An explicit rule for the sequence is an = −(−3)n−1.
40
20
2
4n
The sequence is arithmetic, with first term a1 = 5 and
common difference d = 15.
an = an−1 + d
an = an−1 + 15
So, a recursive rule for the sequence is a1 = 5, an = an−1 + 15.
an = a1 + (n − 1)d
an = 5 + (n − 1)(15)
an = 5 + n(15) − 1(15)
an = 5 + 15n − 15
an = 15n − 10
An explicit rule for the sequence is an = 15n − 10.
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Algebra 1
Worked-Out Solutions
363
Chapter 6
38.
20
41. Find the difference of each term and the one before it.
an
a2 − a1 = 4 − 2 = 2
a3 − a2 = 2 − 4 = −2
10
2
4n
−10
−20
The sequence is arithmetic, with first term a1 = 19 and
common difference d = −13.
an = an−1 + d
an = an−1 − 13
So, a recursive rule for the sequence is a1 = 19, an = an−1 − 13.
an = a1 + (n − 1)d
an = 19 + (n − 1)(−13)
an = 19 + n(−13) − 1(−13)
an = 19 − 13n + 13
an = −13n + 32
An explicit rule for the sequence is an = −13n + 32.
39. Find the sum of each pair of consecutive terms.
a1 + a2 = 1 + 3 = 4
a2 + a3 = 3 + 4 = 7
a3 + a4 = 4 + 7 = 11
Beginning with the third term, each term is the sum of the
two previous terms. So, use the recursive equation
an = an−2 + an−1 to find the next two terms.
a4 − a3 = −2 − 2 = −4
a5 − a4 = −4 − (−2) = −2
Beginning with the third term, each term is the difference of
the previous term and the one before that. So, use the recursive
equation an = an−1 − an−2 to find the next two terms.
a7 = a6 − a5 = −2 − (−4) = 2
a8 = a7 − a6 = 2 − (−2) = 4
So, a recursive rule for the sequence is a1 = 2, a2 = 4,
an = an−1 − an−2, and the next two terms are 2 and 4.
42. Find the sum of each pair of consecutive terms.
a1 + a2 = 6 + 1 = 7
a2 + a3 = 1 + 7 = 8
a3 + a4 = 7 + 8 = 15
a4 + a5 = 8 + 15 = 23
Beginning with the third term, each term is the sum of the
two previous terms. So, use the recursive equation
an = an−2 + an−1 to find the next two terms.
a7 = a5 + a6 = 15 + 23 = 38
a8 = a9 + a10 = 23 + 38 = 61
So, a recursive rule for the sequence is a1 = 6, a2 = 1,
an = an−2 + an−1, and the next two terms are 38 and 61.
43. Find the product of each pair of consecutive terms.
a1(a2) = 1(3) = 3
a6 = a4 + a5 = 7 + 11 = 18
a2(a3) = 3(3) = 9
a7 = a5 + a6 = 11 + 18 = 29
a3(a4) = 3(9) = 27
So, a recursive rule for the sequence is a1 = 1, a2 = 3,
an = an−2 + an−1, and the next two terms are 18 and 29.
Beginning with the third term, each term is the sum of
the two previous terms. So, use the recursive equation
an = (an−2)(an−1) to find the next two terms.
40. Find the difference of each pair of consecutive terms.
a1 − a2 = 10 − 9 = 1
a2 − a3 = 9 − 1 = 8
a3 − a4 = 1 − 8 = −7
a4 − a5 = 8 − (−7) = 15
a6 = a4(a5) = 9(27) = 243
a7 = a5(a6) = 27(243) = 6561
So, a recursive rule for the sequence is a1 = 1, a2 = 3,
an = (an−2)(an−1), and the next two terms are 243 and 6561.
Beginning with the third term, each term is the difference
of the two previous terms. So, use the recursive equation
an = an−2 − an−1 to find the next two terms.
a7 = a5 − a6 = −7 − 15 = −22
a8 = a6 − a7 = 15 − (−22) = 37
So, a recursive rule for the sequence is a1 = 10, a2 = 9,
an = an−2 − an−1, and the next two terms are −22 and 37.
364
Algebra 1
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Chapter 6
44. Find the quotient of each pair of consecutive terms.
64
a
—1 = — = 4
a2 16
16
a
—2 = — = 4
4
a3
4
a3 —
—= =1
a4 4
Beginning with the third term, each term is the quotient of
the two previous terms.
an−2
So, use the recursive equation an = —
to find the next two
an−1
terms.
a
4
a6 = —4 = — = 4
a5 1
a
1
a7 = —5 = —
a6 4
So, a recursive rule for the sequence is a1 = 64, a2 = 16,
an−2
1
, and the next two terms are 4 and —.
an = —
an−1
4
45. The common difference is −12, not 12. The sequence is
arithmetic, with first term a1 = 6 and common difference
d = −12.
an = a1 + (n − 1)d
an = 6 + (n − 1)(−12)
an = 6 + n(−12) − 1(−12)
an = 6 − 12n + 12
an = −12n + 18
An explicit rule for the sequence is an = −12n + 18.
46. The difference between the fourth term, 10, and the third
term, 6, is 4, not 2. The sequence does not have a common
difference for every pair of consecutive terms. So, it is not
arithmetic.
48. The function represents a geometric sequence, with first term
a1 = −1 and common ratio r = 6.
an = a1r n−1
an = −1(6)n−1
Use the explicit rule an = −(6)n−1 to find the 2nd, 5th, and
10th terms of the sequence.
a2 = −(6)2−1 = −61 = −6
a5 = −(6)5−1 = −64 = −1296
a10 = −(6)10−1 = −69 = −10,077,696
49. The function represents a geometric sequence, with first term
a1 = 8 and common ratio r = −1.
an = a1r n−1
an = 8(−1)n−1
Use the explicit rule an = 8(−1)n−1 to find the 2nd, 5th, and
10th terms of the sequence.
a2 = 8(−1)2−1 = 8(−1)1 = 8(−1) = −8
a5 = 8(−1)5−1 = 8(−1) 4 = 8(1) = 8
a10 = 8(−1)10−1 = 8(−1) 9 = 8(−1) = −8
50. a1 = 4
a2 = 5
a3 = a1 + a2 = 4 + 5 = 9
a4 = a2 + a3 = 5 + 9 = 14
a5 = a3 + a4 = 9 + 14 = 23
a6 = a4 + a5 = 14 + 23 = 37
a7 = a5 + a6 = 23 + 37 = 60
a8 = a6 + a7 = 37 + 60 = 97
a9 = a7 + a8 = 60 + 97 = 157
Find the sum of each pair of consecutive terms.
a10 = a8 + a9 = 97 + 157 = 254
a1 + a2 = 2 + 4 = 6
So, the 2nd term is 5, the 5th term is 23, and the 10th term
is 254.
a2 + a3 = 4 + 6 = 10
a3 + a4 = 6 + 10 = 16
Beginning with the third term, each term is the sum of the
two previous terms. So, a recursive rule for the sequence is
a1 = 2, a2 = 4, an = an−2 + an−1.
47. The function represents an arithmetic sequence, with first
term a1 = 3 and common difference d = 7.
an = a1 + (n − 1)d
an = 3 + (n − 1)7
an = 3 + n(7) − 1(7)
an = 3 + 7n − 7
an = 7n − 4
Use the explicit rule an = 7n − 4 to find the 2nd, 5th, and
10th terms of the sequence.
51. a1 = 10
a2 = 15
a3 = a2 − a1 = 15 − 10 = 5
a4 = a3 − a2 = 5 − 15 = −10
a5 = a4 − a3 = −10 − 5 = −15
a6 = a5 − a4 = −15 − (−10) = −5
a7 = a6 − a5 = −5 − (−15) = 10
a8 = a7 − a6 = 10 − (−5) = 15
a9 = a8 − a7 = 15 − 10 = 5
a10 = a9 − a8 = 5 − 15 = −10
So, the 2nd term is 15, the 5th term is −15, and the 10th
term is −10.
a2 = 7(2) − 4 = 10
a5 = 7(5) − 4 = 31
a10 = 7(10) − 4 = 66
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Algebra 1
Worked-Out Solutions
365
Chapter 6
52. a. Starting with the smallest bone, the first term is a1 = 2.5,
c. The spreadsheet generates the following recursively
defined sequence.
and the second term is a2 = 3.5. Find the sum of each pair
of consecutive terms.
4
a1 + a2 = 2.5 + 3.5 = 6
7
a2 + a3 = 3.5 + 6 = 9.5
3
Beginning with the third term, each term is the sum of
the two previous terms. So, use the recursive equation
an = an−2 + an−1 to find the next two terms.
−4
a6 = a4 + a5 = 7 + 11 = 18
−3
−7
a7 = a5 + a6 = 11 + 18 = 29
4
So, a recursive rule for the sequence is a1 = 2.5
centimeters, a2 = 3.5 centimeters, and an = (an−2 + an−1)
centimeters.
7
3
−4
b. Sample answer: A sequence is 2.7 cm, 2.4 cm, 2.5 cm,
8.5 cm. This sequence does not follow a pattern.
54. a.
53. a. The spreadsheet generates the following arithmetic
sequence.
Square, n
1
2
3
4
5
6
Side length, an
1
1
2
3
5
8
The sequence is 1, 1, 2, 3, 5, 8.
3
b. This is the Fibonacci sequence, and the next term is
5
5 + 8 = 13.
7
c.
9
5
11
6
13
21
15
3
17
4
19
21
b. The spreadsheet generates the following geometric
sequence.
7
3
12
48
192
768
3072
12,288
49,152
196,608
786,432
366
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Chapter 6
Maintaining Mathematical Proficiency
55. a1 = 5
a2 = 3(5) + 4 = 19
60. 5x + 12x = 17x
a3 = 3(19) + 4 = 61
61. 9 − 6y − 14 = −6y + 9 − 14
a4 = 3(61) + 4 = 187
= −6y − 5
a5 = 3(187) + 4 = 565
The first five terms of the sequence are 5, 19, 61, 187, and 565.
62. 2d − 7 − 8d = 2d − 8d − 7
= −6d − 7
Find the difference between each pair of consecutive terms.
5
⤻
19
⤻
61
⤻
187
⤻
565
63. 3 − 3m + 11m = 11m − 3m + 3
= 8m + 3
19 − 5 = 14 61 − 19 = 42 187 − 61 = 126 565 − 187 = 378
Find the ratio between each pair of consecutive terms.
5
⤻
⤻
—
—
19
19
5
61
61
19
⤻
⤻
187
187
61
64.
565
y − 6 = 3(x − 2)
565
—
187
—
y − 6 = 3(x) − 3(2)
The sequence has neither a common difference nor a
common ratio. So, it is neither arithmetic nor geometric.
y − 6 = 3x − 6
+6
the two terms above it in the previous row. So, a recursive rule
that defines the mth term in the nth row is a1,n = 1, an, n = 1,
am, n = am−1, n−1 + am, n−1.
57. a. Substituting n − 1 for n in the explicit rule that defines an
arithmetic sequence results in this expression.
Write the equation.
= a1 + [(n − 1) + 0]d
Identity Property of
Addition
= a1 + [(n − 1) + −1 + 1]d Additive Inverse Property
= a1 + [((n − 1) − 1) + 1]d Associative Property of
Addition
= a1 + [(n − 1) −1]d + d
Distributive Property
= an−1 + d
Substitution Property of
Equality
A function is f (x) = 3x.
1
−4 − 0 −4
−4 − 0
m = — = — = —, or −—
2
6 − (−2)
6+2
8
y − y1 = m(x − x1)
1
y − 0 = −—[ x − (−2) ]
2
1
y = −—(x + 2)
2
1
1
y = −—(x) − —(2)
2
2
1
y = −—x − 1
2
1
A function is f (x) = −—x − 1.
2
5−5
0
5−5
66.
m = — = — = —, or 0
−1 − (−3) −1 + 3 2
y − y1 = m(x − x1)
65.
y − 5 = 0[ x − (−3) ]
58. Find the ratio between each pair of consecutive terms.
−5
5
⤻ ⤻
5
−5
— = −1
−5
5
−5
5
⤻ ⤻
y−5=0
−5
+5+5
−5
5
5
−5
— = −1 — = −1 — = −1
y=5
Your friend is not correct. The sequence is a geometric
sequence, with first term a1 = −5 and a common factor of
−1. So, it can be represented by the recursive rule
a1 = −5, an = −an−1.
59. Find the difference between each pair of consecutive terms.
3
⤻
7
7−3=4
⤻
15
15 − 7 = 8
⤻
31
31 − 15 = 16
⤻
63
63 − 31 = 32
Each difference is twice the previous difference. So, a
recursive rule for the sequence is a1 = 3, an = an−1 + 2n.
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+6
y = 3x
56. Each row begins and ends with 1, and each term is the sum of
b. an = a1 + (n − 1)d
−3 − 6 −9
m = — = —, or 3
−1 − 2 −3
y − y1 = m(x − x1)
A function is f (x) = 5.
67.
−15 − (−1) −15 + 1 −14
m = —— = — = —, or 2
−4 − 3
−4 − 3
−7
y − y1 = m(x − x1)
y − (−1) = 2(x − 3)
y + 1 = 2(x) − 2(3)
y + 1 = 2x − 6
−1
−1
y = 2x − 7
A function is f (x) = 2x − 7.
Algebra 1
Worked-Out Solutions
367
Chapter 6
6.5 –6.7 What Did You Learn? (p. 347)
1. Sample answer: Because the growth factor ended in the
hundredths place, the answer should be rounded to the
hundredths place.
2. Sample answer: Use the function from part (a) to calculate
9.
−2
x
()
1
−4 __
4
1
−4 —
4
−4
−5
1
x4
2. —7 = x4−7 = x−3 = —3
x
x
3.
(x0y2)3
= (1
⋅
y2)3
f(x) = −4(
2x2
5y
3—
−1
()
1
−4 —
4
1
−—
4
2
4 x
−60
−80
The domain is all real numbers and the range is y < 0.
10. x
−2
−1
3−2+2 3−1+2
3x+2
f (x)
1
60
(2x2)−2 (5y4)2 52(y4)2 25y4⋅2 25y8
=—
=—=—=—
=—
(5y4)−2 (2x2)2 22(x2)2
4x2⋅2
4x4
0
1
2
30+2
31+2
32+2
9
27
81
3
= y6
⋅2⋅2
()
1
−4 —
4
−40
80
−2
()
)
= y2⋅3
( )
−4
2
= (y2)3
4. —4
−16
−2
= y3+(−5)
= y−2
1
= —2
y
1
−4 —
4
2
1
1 x
4
Chapter 6 Review (pp. 348–350)
⋅y
1
0
y
measured in centimeters, so a 30-centimeter ruler is an
appropriate tool to use.
1. y3
0
−1
()
1
−4 —
4
−64
f (x)
the number of squares for the tenth figure in part (b) without
determining the number of squares in the fourth through
ninth figures.
3. Sample answer: Parts of a hand can be conveniently
−1
−2
()
x
y
40
f(x) = 3x+2 20
3—
5. √ 8 = √ 2
=2
5—
5 ———
6. √ −243 = √ (−3)(−3)(−3)(−3)(−3)
= −3
7. 6253/4 = (6251/4)3
4— 3
= ( √625 )
4 —— 3
= ( √5 ⋅ 5 ⋅ 5 ⋅ 5 )
−6
−4
2 x
The domain is all real numbers and the range is y > 0.
11. x
2
3
4
2x−4 − 3 2−2 − 3 2−1 − 3
20 − 3
−2.75
−2
f (x)
= (5)3
x
= 125
2x−4
8. (−25)1/2 is not a real number because there is no real number
−2
−2.5
5
−3
21
18
12
−3
−1
f (x)
that can be multiplied by itself twice to produce −25.
6
22
−3
1
7
23
−3
5
8
24
−3
13
y
f(x) = 2 x−4 − 3
6
2
6
8 x
−6
The domain is all real numbers and the range is y > −3.
368
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Chapter 6
1
2
12. The first term is 2, and the common ratio is —, or 0.5.
f (x) = abx
f (x) =
= (1.068)t
≈ (1.59)t
2(0.5)x
A function represented by the table is f (x) = 2(0.5)x.
16
16. y = (1.06)8t
y
The function is of the form y = a(1 + r)t, where 1 + r > 1.
So, it represents exponential growth. Use the factor 1 + r
to find the rate of growth.
1 + r = 1.59
12
−1
8
1 x
f(x) = 2( 2)
4
−4
−2
2
−1
r = 0.59
So, the function represents exponential growth, and the rate
of growth is 59%.
4 x
17. f (t) = 6(0.84)t−4
(0.84)t
= 6—4
(0.84)
6
≈ —(0.84)t
0.49787
≈ 12.05(0.84)t
The graph of f is a vertical stretch by a factor of 2 of the
graph of g.
+1 ⤻
+1 ⤻
+1
⤻
13.
x
0
1
2
3
y
3
6
12
24
⤻⤻
⤻
×2 ×2 ×2
As x increases by 1, y is multiplied by 2. So, the table
represents an exponential growth function.
+1 ⤻
+1 ⤻
+1
⤻
14.
x
1
2
3
4
y
162
108
72
48
⤻
⤻
⤻
2
2
2
×— ×— ×—
3
3
3
2
As x increases by 1, y is multiplied by —. So, the table
3
represents an exponential decay function.
15. f (t) = 4(1.25)t+3
⋅ (1.25)
f (t) = 4(1.25) ⋅ (1.953125)
f (t) = 4(1.25)t
3
t
f (t) = 7.8125(1.25)t
The function is of the form y = a(1 + r)t, where 1 + r > 1.
So, it represents exponential growth. Use the growth factor
1 + r to find the rate of growth.
1 + r = 1.25
−1
−1
The function is of the form y = a(1 − r)t, where 1 − r < 1.
So, it represents exponential decay. Use the decay factor
1 − r to find the rate of decay.
1−r=
−1
0.84
−1
− r = −0.16
−0.16
−r
—=—
−1
−1
r = 0.16
So, the function represents exponential decay, and the rate of
decay is 16%.
(
r
n
18. a. y = P 1 + —
(
nt
)
)
0.05 4t
y = 750 1 + —
4
y = 750(1 + 0.0125)4t
y = 750(1.0125)4t
A function that represents the balance after t years is
y = 750(1.0125)4t.
b. y = 750(1.0125)4(4)
≈ 750(1.21989)
≈ 914.92
The balance of the account is $914.92 after 4 years.
r = 0.25
So, the function represents exponential growth, and the rate
of growth is 25%.
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Algebra 1
Worked-Out Solutions
369
Chapter 6
19. a. The initial value is $1500, and the rate of decay is
21. 3x−2 = 1
14%, or 0.14.
Use the fact that any number to the zero power is 1. So x − 2
must equal 0.
y = a(1 − r)t
= 1500(1 − 0.14)t
x−2= 0
= 1500(0.86)t
+2 +2
1
12
exponents to rewrite the function in a form that reveals the
monthly rate of decay.
b. Use the fact that t = —(12t) and the properties of
x=2
The solution is x = 2.
22. −4 = 64x−3
y = 1500(0.86)t
Graph y = 64x−3 and y = −4.
= 1500(0.86)(1/12)(12t)
= 1500(0.861/12)12t
5
≈ 1500(0.9875)12t
Use the decay factor 1 − r ≈ 0.9875 to find the rate of
decay.
−5
5
1 − r ≈ 0.98751
−1
−5
−1
The graphs do not intersect. So, the equation has no solution.
− r ≈ − 0.0125
−r −0.01249
—≈—
−1
−1
r ≈ 0.01249
()
1
3
23. —
2x+3
()
1
Graph y = —
3
So, the monthly percent decrease is about 1.2%.
c.
Year, t
0
Balance
(dollars), y
2
4
6
−6
Balance (dollars)
t
24.
( 161 )
( 41 )
—
—2
6
3x
3x
= 642(x+8)
= (43)2(x+8)
(4−2)3x = 43⋅2(x+8)
4−2(3x) = 46(x+8)
0 1 2 3 4 5 6 7 8 t
From the graph, you can see that the value of the TV is
about $950 after 3 years.
5x = 53x−2
x = 3x − 2
− 3x
Intersection
X=-2.232487 Y=5
−2
The solution is x ≈ −2.23.
Year
20.
and y = 5.
8
1500 1109.4 820.51 606.85 448.83
y = 1500(0.86)
2x+3
6
TV Value
y
1600
1400
1200
1000
800
600
400
200
0
=5
− 3x
− 2x = −2
−2x −2
—=—
−2
−2
x=1
−2(3x) = 6(x + 8)
−6x = 6(x) + 6(8)
−6x = 6x + 48
− 6x
− 6x
−12x = 48
48
−12x
—=—
−12
−12
x = −4
The solution is x = −4.
The solution is x = 1.
370
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Chapter 6
28. Position
272x+2 = 81x+4
25.
(33)2x+2 = (34)x+4
Term
33(2x+2) = 34(x+4)
4
5
375
−75
15
−3
—
3
5
6
3
−—
25
7
3
125
—
( ) ( )( )( )( )( )
6x + 6 = 4x + 16
− 4x
2x + 6 = 16
−6
2x = 10
10
2x —
—=
2
2
x=5
450
an
(1, 375)
300
150
The solution is x = 5.
26.
3
1
1
1
1
1
1
× −— × −— × −— × −— × −— × −—
5
5
5
5
5
5
1
Each term is −— times the previous term. So, the sequence
5
1
is geometric with a common ratio of −—, and the next three
5
3
3
3
terms are —, −—, and —.
5 25
125
3(2x) + 3(2) = 4(x) + 4(4)
−6
2
⤻⤻⤻⤻⤻ ⤻
3(2x + 2) = 4(x + 4)
− 4x
1
(5, 53 )
(4, −3)
(3, 15)
2
Position
1
2
3
4
5
Term
3
12
48
192
768
6
−150
7
3072 12,288
⤻
⤻⤻⤻⤻
⤻
×4 ×4 ×4 ×4
×4 ×4
29.
Each term is 4 times the previous term. So, the sequence is
geometric with a common ratio of 4, and the next three terms
are 768, 3072, and 12,288.
(7, 1235)
)
(
3
6, −25
4
8n
6
(2, −75)
Position, n
1
2
3
4
Term, an
1
4
16
64
⤻
⤻⤻
×4 ×4 ×4
The first term is a1 = 1 and the common ratio is r = 4.
an = a1r n−1
an
an = 1(4)n−1
16,000
an = 4n−1
(7, 12,288)
12,000
Use the equation to find the 9th term.
8000
(3, 48)
(2, 12)
4000
(1, 3)
2
(4, 192)
a9 = 49−1
(6, 3072)
(5, 768)
a9 = 48
4
a9 = 65,536
8n
6
An equation for the nth term is an = 4n−1 and a9 = 65,536.
27.
Position
1
Term
9
2
3
4
−18 27 −36
⤻ ⤻⤻
3
Position
1
Term
9
4
×−2 ×−—2 ×−—3
2
3
4
−18 27 −36
⤻ ⤻⤻
30.
Position, n
1
2
3
4
Term, an
5
−10
20
−40
+(−27) +45 +(−63)
The sequence has neither a common ratio nor a common
difference. So, it is neither arithmetic nor geometric.
⤻ ⤻⤻
×(−2) ×(−2) ×(−2)
The first term is a1 = 5 and the common ratio is r = −2.
an = a1rn−1
an = 5(−2)n−1
Use the equation to find the 9th term.
a9 = 5(−2)9−1
= 5(−2)8
= 5(256)
= 1280
An equation for the nth term is an = 5(−2)n−1 and
a9 = 1280.
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Algebra 1
Worked-Out Solutions
371
Chapter 6
31.
Position, n
Term, an
1
2
3
4
486
162
54
18
33. a1 = −4
a2 = −3a1 = −3(−4) = 12
⤻
⤻
⤻
1
1
1
×— ×—
×—
3
3
3
1
The first term is a1 = 486 and the common ratio is —.
3
an = a1rn−1
1 n−1
an = 486 —
3
Use the equation to find the 9th term.
()
()
1
= 486( )
3
1
= 486( )
3
1
= 486(
6561 )
1
a9 = 486 —
3
—
a3 = −3a2 = −3(12) = −36
a4 = −3a3 = −3(−36) = 108
a5 = −3a4 = −3(108) = −324
a6 = −3a5 = −3(−324) = 972
The first six terms are −4, 12, −36, 108, −324, and 972.
an
1200
9−1
800
8
400
—8
2
4
8n
6
−400
—
34. a1 = 32
2
=—
27
()
1
An equation for the nth term is an = 486 —
3
n−1
32. a1 = 4
a2 = a1 + 5 = 4 + 5 = 9
a3 = a2 + 5 = 9 + 5 = 14
a4 = a3 + 5 = 14 + 5 = 19
a5 = a4 + 5 = 19 + 5 = 24
a6 = a5 + 5 = 24 + 5 = 29
The first six terms are 4, 9, 14, 19, 24, and 29.
an
2
and a9 = —.
27
1
1
a2 = — a1 = — (32) = 8
4
4
1
1
a3 = — a2 = — (8) = 2
4
4
1
1
1
a4 = — a3 = — (2) = —
4
4
2
()
1
1 1
1
a = a = ( )=
4
4 8
32
1
1 1
1
a5 = — a4 = — — = —
4
4 2
8
6
— 5
— —
—
1
1 1
The first six terms are 32, 8, 2, — , — , and —.
2 8
32
an
32
32
24
24
16
16
8
8
2
4
6
8n
2
4
6
8n
35. Find the difference between each pair of consecutive terms.
3
8
8−3=5
13
13 − 8 = 5
18 − 13 = 5
18
23
23 − 18 = 5
The sequence is arithmetic, with first term a1 = 3 and
common difference d = 5.
an = an−1 + d
an = an−1 + 5
So, a recursive rule for the sequence is a1 = 3, an = an−1 + 5.
372
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Worked-Out Solutions
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Chapter 6
4. z−2
36. Find the ratio between each pair of consecutive terms.
3
6
12
24
48
24
48
12
6
—=2
—=2
—=2
6
12
24
3
The sequence is geometric, with first term a1 = 3 and
common ratio r = 2.
—=2
an = r
⋅a
4
b−5
ab
1
a1 + a2 = 7 + 6 = 13
a2 + a3 = 6 + 13 = 19
a3 + a4 = 13 + 19 = 32
Beginning with the third term, each term is the sum of the
two previous terms. So, a recursive rule for the sequence is
a1 = 7, a2 = 6, an = an−2 + an−1.
an
⋅b
= b8−5
= b3
−3
2c4
5
37. Find the sum of each pair of consecutive terms.
= z2
b8
6. —
So, a recursive rule for the sequence is a1 = 3, an = 2an−1.
= z−2+4
5. —
= —5
0 −8
( )
n−1
an = 2an−1
38.
⋅z
(2c4)−3
53
125
125
125
=—
=—
= —= —= —
5−3
(2c4)3 23(c4)3 8c4⋅3 8c12
7. The initial amount is 42,500 and the rate of growth is 3%, or
0.03.
y = a(1 + r)x
y = 42,500(1 + 0.03)x
y = 42,500(1.03)x
A function that represents the situation is y = 42,500(1.03)x.
x
0
2
42,500(1.03)0
42,500(1.03)2
y
42,500
45088.25
x
4
6
42,500(1.03)4
42,500(1.03)6
47834.12
50,747.22
42,500(1.03) x
1000
750
42,500(1.03) x
500
y
250
The sequence is geometric, with first term a1 = 8 and
common ratio r = 5.
an = r
⋅a
n−1
an = 5an−1
So, a recursive rule for the sequence is a1 = 8, an = 5an−1.
an = a1r n−1
an = 8(5)n−1
Salary (thousand dollars)
Salary
4n
2
y
60.0
52.5
45.0
37.5
30.0
22.5
15.0
7.5
0
y = 42,500(1.03)x
0 1 2 3 4 5 6 7 8 x
Year
An explicit rule for the sequence is an = 8(5)n−1 .
Chapter 6 Test (p. 351)
4—
⋅2⋅2⋅2
4 ——
1. −√ 16 = −√ 2
= −(2)
= −2
6—
2. 7291/6 = √ 729
⋅3⋅3⋅3⋅3⋅3
6 ——
= √3
=3
3. (−32)7/5 = ( (−32)1/5 )7
= ( √ −32 )
5— 7
=
———
( √5 (−2)(−2)(−2)(−2)(−2) )7
= (−2)7
= −128
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Algebra 1
Worked-Out Solutions
373
Chapter 6
+1
⤻
8. The initial amount is 500, the annual interest rate is 6.5%, or
0.065, and the number of times n that interest is compounded
per year is 1.
r nx
y=P 1+—
n
)
(
(
0.065
y = 500 1 + —
1
y=
10.
)
x
0
2
4
500(1.065)0
500(1.065)2
500(1.065)4
500
567.11
643.23
x
500(1.065) x
6
8
500(1.065)6
500(1.065)8
729.57
827.50
y
Amount (dollars)
100
25
⤻ ⤻
4
⤻
6.25
n−1
()
1
So, an explicit rule for the sequence is an = 400 —
4
an = r an−1
⋅
n−1
.
1
an = — an−1
4
So, a recursive rule for the sequence is a1 = 400,
1
an = — an−1.
4
1
128
1
Check: 2x = —
128
? 1
2−7 = —
128
1 ? 1
—7 = —
128
2
11. 2x = —
y = 500(1.065)x
1
2x = —7
2
2x = 2−7
0 1 2 3 4 5 6 7 8 x
1
128
+1
⤻
+1
⤻
12.
n
1
2
3
4
an
−6
8
22
36
+14
⤻
+14
⤻
+14
The sequence is arithmetic, with first term a1 = −6 and
common difference d = 14.
an = a1 + (n − 1)d
an = −6 + (n − 1)(14)
an = −6 + n(14) − 1(14)
an = −6 + 14n − 14
an = 14n − 20
So, an explicit rule for the sequence is an = 14n − 20.
1
128
—=—✓
x = −7
+1
⤻
⤻
3
()
Year
9.
400
1
an = 400 —
4
Deposit Account
y
800
700
600
500
400
300
200
100
0
an
2
1
1
1
×—
×—
×—
4
4
4
The sequence is geometric, with first term a1 = 400 and
1
common ratio r = —.
4
an = a1rn−1
A function that represents the situation is y = 500(1.065)x.
y
1
1x
500(1.065)x
500(1.065) x
n
+1
⤻
+1
⤻
256x+2 = 163x−1
Check: 256x+2 = 163x−1
?
=
2565+2 = 163(5)−1
?
44(x +2) = 42(3x−1)
2567 = 1614
?
4(x + 2) = 2(3x − 1)
2567 = 162⋅7
?
4(x) + 4(2) = 2(3x) − 2(1)
2567 = (162)7
(44)x +2
(42)3x−1
4x + 8 = 6x − 2
− 4x
2567 = 2567 ✓
− 4x
8 = 2x − 2
+2
+2
10 = 2x
10 2x
—=—
2
2
5=x
an = an−1 + d
an = an−1 + 14
So, a recursive rule for the sequence is a1 = −6,
an = an−1 + 14.
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Chapter 6
13.
x
2(6)x
−1
0
1
2
3
2(6)−1
2(6)0
2(6)1
2(6)2
2(6)3
1
—
3
2
12
72
432
y
400
an = a1r n−1
an = 1(1.2)n−1
So, an explicit rule for the sequence is an = 1.2n−1.
an = r
⋅a
n−1
an = 1.2an−1
200
So, a recursive rule for the sequence is a1 = 1,
an = 1.2an−1.
100
f(x) = 2(6)x
−2
common ratio r = 1 + 0.20 = 1.2.
an = 1.2n−1
y
300
−4
18. a. The sequence is geometric, with first term a1 = 1 and
b. In order to solve 3 ≈ 1.2n−1, graph y = 3 and y = 1.2n−1,
4 x
2
and find the point of intersection.
6
The graph of f is a vertical stretch by a factor of 2 of the
graph of g.
From the graph, you can see that the domain is all real
numbers and the range is y > 0.
−4
5a
5
The solution is n ≈ 7.03. So, you run approximately
3 kilometers on day 7.
14. —b = 5−3
5a
By the Quotient of Powers Property, —b = 5a−b. So, the
5
difference a − b must equal −3, and in order for the
difference to be a negative number, it must be that a < b.
15. 9a
⋅
9−b = 1
Chapter 6 Standards Assessment (pp. 352–353)
x5/3
⋅ x ⋅ √x
x ⋅x
−1
3—
⋅
17. a. The initial amount is 1 atmosphere. The decay factor is
0.99988. The decay rate is 1 − 0.99988 = 0.00012, or
0.012%.
b.
5/3
−1
1/3
2
= x(5/3)−1+(1/3)+2
= x3
16. If the sequence is arithmetic, then the common difference is
⋅
⋅ ⋅
⋅
x ⋅x ⋅x ⋅x
= ——
1
x5/3 x−1 x1/3
x 1
1. ——
= ——
−2
−2
0
Any number to the 0 power must be 1, and by the Product of
Powers Property, 9a 9−b = 9a+(−b). So, the sum a + (−b)
must equal 0, which means that a = b.
d = a2 − a1 = −12 − 3 = −15. So, a3 = a2 + d = −12 +
−15 = −27. If the sequence is geometric, then the common
−12
a
ratio is r = —2 = — = −4. So, b3 = r a2 = −4(−12) =
a1
3
48. Then, a3 − b3 = −27 − 48 = −75.
8
Intersection
X=7.0256851 Y=3
−2
So, the exponent of x is 3.
2. f (−1) = 2 = 21
f (−2) = 4 = 22
f (−3) = 8 = 23
So, following the pattern, f (−7) = 27 = 128.
1.25
0
6000
0
At an altitude of 5000 feet, the atmospheric pressure is
about 0.549 atmosphere.
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Algebra 1
Worked-Out Solutions
375
Chapter 6
3. a. Sample answer: Because 3(3x + y = 12) is 9x + 3y = 36,
which is equivalent to 3y + 9x = 36, the equations
3x + y = 12 and 3y + 9x = 36 form a linear system that
has infinitely many solutions.
b. Sample answer:
Step 2:
Multiply by −2.
−6x − 2y = −24
3x + 2y = 12
3x + 2y = 12
−3x + 0 = −12
Step 3:
−3x = −12
−3x −12
—=—
−3
−3
x=4
Step 4:
3x + y = 12
3(4) + y = 12
12 + y = 12
− 12
− 12
y=0
So, the equations 3x + y = 12 and 3x + 2y = 12 form a
linear system that has one solution (4, 0).
c. Sample answer: Try 3x + y = 12 and 6x + 2y = 6.
Solve by elimination.
an = an−1 + 10
an = a1 + (n − 1)d
an = −3 + (n − 1)(10)
an = −3 + n(10) − 1(10)
an = −3 + 10n − 10
an = 10n − 13
So, an explicit rule for the sequence is an = 10n − 13.
6. B; When the equation is rewritten in slope-intercept form,
it is y = −14.8t + 870. So, m = −14.8 is the rate of the
descent, and b = 870 is the initial height.
7. f (x) = 3x − 2
f (x) = −2x + 4
10 = 3x − 2
10 = −2x + 4
+2
+2
−4
Multiply by −2.
6x + 2y = 6
−6x − 2y = −24
6x + 2y =
6
0 = −18
Because 0 = −18 is never true, the equations 3x + y = 12
and 6x + 2y = 6 form a linear system that has no solution.
y − 2x ≤ 4
6x − 3y < −12
y − 2x + 2x ≤ 4 + 2x
6x − 6x − 3y < −12 − 6x
y ≤ 2x + 4
−3y < −6x −12
−3y
3
−6x − 12
−3
— < —
y > 2x + 4
−4
12 = 3x
6 = −2x
12 3x
3
3
4=x
−2x
6
−2
−2
−3 = x
1
f (x) = —x − 6
2
1
10 = —x − 6
2
+6
+ 6
—=—
—=—
f (x) = −3x + 5
Step 1:
4.
an = an−1 + d
a1 = −3, an = an−1 + 10.
Step 1:
3x + y = 12
a1 = 7 − 10 = −3 and common difference d = 10.
So, a recursive rule for the sequence is
Solve by elimination:
3x + y = 12
5. The sequence is arithmetic, with first term
10 = −3x + 5
−5
−5
1
16 = —x
2
5 = −3x
−3x
5
−3
−3
5
−— = x
3
—=—
2
2
3
—
⋅
3
6 = —x
2
2 3
6 = — —x
3 2
4=x
⋅
f (x) = 4x + 14
10 = 4x + 14
− 14
− 14
−4 = 4x
⋅ 16 = 2 ⋅ —2x
1
32 = x
3
f (x) = — x + 4
2
3
10 = —x + 4
2
− 4
− 4
−4
4
4x
4
—=—
−1 = x
The functions whose x-value is an integer when f (x) = 10
3
are f (x) = 3x − 2, f (x) = −2x + 4, f (x) = — x + 4,
2
1
f (x) = — x − 6, and f (x) = 4x + 14.
2
In order for the system to have no solution, the last step of
the equation should be y > 2x + 4. Because the symbol
has to be reversed when each side is divided by −3,
the original equation must be 6x − 3y < −12 when the
system has no solution.
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Chapter 6
8. Functions of the form y = a(1 + r)x, where a > 0 and r > 0,
are exponential growth functions. The exponential growth
1
functions in the list are f (x) = — (3)x and f (x) = 4(1.6)x/10.
2
Functions of the form y = a(1 − r)x, where a > 0 and
0 < r < 1 are exponential decay functions. The only
x
1
exponential decay function in the list is f (x) = 3 — .
6
Also, the function f (x) = −2(8)x is of the form y = abx,
which means it is exponential, but it is neither an exponential
growth nor an exponential decay function.
()
Linear functions have variables with an exponent of 0 or 1.
So, the linear functions in the list are f (x) = 15 − x,
4—
f (x) = −3(4x + 1 − x), and f (x) = √16 + 2x.
The functions that are neither exponential nor linear are
f (x) = 6x2 + 9 and f (x) = x(18 − x), because one has an
exponent of 2 or an x, and the other has an exponent of 2
when you use Distributive Property.
9. The graph shown is a horizontal translation 1 unit right and a
vertical translation 3 units up of the graph of f.
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Worked-Out Solutions
377