7.1 SOLUTIONS
CHAPTER SEVEN
Solutions for Section 7.1
EXERCISES
1. Since A = 1 · x2 , the exponent is 2 and the constant of proportionality is 1.
2. Since P = 4x1 , the exponent is 1 and the constant of proportionality is 4.
√
3. Since x = 3 V = 1 · V 1/3 , the exponent is 1/3 and the constant of proportionality is 1.
4. The power of r is 1 and the constant of proportionality is 2π.
5. The power of r is 2 and the constant of proportionality is 4π.
6. (a) The coefficient is 3 and the exponent is 2.
(b) We have
Area = 3w2 = 3 · 52 = 75 cm2 .
7. (a) The coefficient is 15 π and the exponent is 3.
(b) We have
1 3
1
8
πr = π · 23 = π cm3 .
5
5
5
(c) Since we know that the radius is 5 times the height, the radius is 5 · 0.8 = 4 cm, so
Volume =
Volume =
1
64
1 3
πr = π · 43 =
π cm3 .
5
5
5
8. (a) The coefficient is 4.9 and the exponent is 2.
(b) After 2 seconds, the ball is at
Depth = 4.9t2 = 4.9 · 22 = 19.6 meters.
(c) After 4 seconds, the ball is at
Thus, the hole is 78.4 meters deep.
Depth = 4.9t2 = 4.9 · 42 = 78.4 meters.
9. p > 1
10. 0 < p < 1
11. p = 1
12. p < 0
13. p > 1
14. (a) We see that y is proportional to the second power of x.
(b) We have
x
1
10
100
1000
y = 2x2
2
200
20,000
2,000,000
(c) We see in the table that y increases as x gets larger.
215
216
Chapter Seven /SOLUTIONS
15. (a) We see that y is proportional to the one-half power of x.
(b) We have
x
√
y=3 x
1
10
100
1000
3
9.49
30
94.87
(c) We see in the table that y increases as x gets larger.
16. (a) We see that y is inversely proportional to x.
(b) We have
x
1
10
100
1000
y = 1/x
1
0.1
0.01
0.001
(c) We see in the table that y decreases as x gets larger.
17. (a) We see that y is inversely proportional to the second power of x.
(b) We have
x
1
10
100
1000
y = 5/x2
5
0.05
0.0005
0.000005
(c) We see in the table that y decreases as x gets larger.
PROBLEMS
18. (a) The graph is symmetric about the y-axis, so p is even.
(b) Since the values of y get small as x gets large, p is negative.
19. (a) The graph crosses the x-axis, so p must be odd.
(b) The function is defined at x = 0, so p must be positive.
20. (a) p is odd, because y changes sign when x does.
(b) p is negative, because the values of f get small when x gets large.
21. As x increases, x4 increases. So if a > b, then a4 > b4 .
22. As x increases, x1/4 increases. So if a > b, then a1/4 > b1/4 .
23. As x increases, x−4 decreases. So if a > b, then a−4 < b−4 .
24. As x increases, x−1/4 decreases. So if a > b, then a−1/4 < b−1/4 .
25. (a) The graph of y = x5 is above the graph of y = x3 when x is greater than 1, and when x is between −1 and 0. It is
below when x is between 0 and 1, and when x is less than −1.
(b) We have x5 > x3 when x > 1 and −1 < x < 0, and x5 < x3 when 0 < x < 1 and when x < −1.
26. (a) The graph of y = x1/2 is above the graph of y = x1/4 when x is greater than 1, and below when x is between 0 and
1.
(b) We have x1/2 > x1/4 when x > 1 and x1/2 < x1/4 when 0 < x < 1.
27. If one graph is above another then the y-values on the first are greater than the corresponding y-values on the second. So
x4 > x2 for x > 1 and x4 < x2 for 0 < x < 1.
28. In Example 1 we saw that the braking distance of an Alfa Romeo going at v mph is
f (v) = 0.036v 2 .
We are told that h(v) = 1.3f (v), so
h(v) = 1.3 · 0.036v 2 = 0.047v 2 .
7.1 SOLUTIONS
217
29. (a) If the job pays $4 an hour, the student has to work 2400/4 = 600 hours. If the job pays $10 an hour, the student has
to work 2400/10 = 240 hours.
(b) As the hourly wage goes up, the number of hours the student has to work goes down. This makes sense algebraically
because w is in the denominator, so as the denominator increases, the value of h decreases. In practical terms, if the
wage increases, then the student has to work fewer hours to earn $2400.
(c) Since
2400
,
h=
w
we have
2400
,
w=
h
So w is inversely proportional to h.
30. When t = 3, we have D = 16(32 ) = 144 feet. When t = 5, we have D = 16(52 ) = 400 feet. So the distance is larger
after five seconds. This makes sense algebraically because 5 > 3, so 52 > 32 .
31. (a) We have
0.15
= 0.15p−3/2 .
p3/2
So E is a power function of p with exponent −3/2 and constant of proportionality 0.15.
(b) When p = 0.16 dollars per kilowatt-hour, we have
E=
E=
0.15
= 2.3438 gigawatt-hours per year.
(0.16)3/2
When p = 0.25 dollars per kilowatt-hour, we have
E=
0.15
= 1.2 gigawatt-hours per year.
(0.25)3/2
The electricity consumption decreases as the price increases. Since 0.16 < 0.25, so (0.16)3/2 < (0.25)3/2 . But the
p3/2 is in the denominator, so a larger value of p means a larger denominator which results in a smaller output.
32. The surface area for a mammal of body mass 70 kilograms is larger because 70 > 60, so 702/3 > 602/3 .
33. (a) The radius is graphed in Figure 7.1.
(b) In Figure 7.1, we estimate that a sphere of radius 2 cm has a volume of approximately 33.5 cm3 .
r (cm)
2
√
r = 0.620 3 V
33.5
V (cm3 )
Figure 7.1
34. The graph of x2 · x3 in Figure 7.2 appears to be the same as the graph of x5 , not of x6 . Thus, it appears that x2 · x3 =
x5 = x2+3 .
218
Chapter Seven /SOLUTIONS
1
1
x2 · x3
x5
x6
x
1
−1
1
x
1
−1
−1
x
1
−1
−1
−1
Figure 7.2
35. The graphs of −x4 and (−x)4 in Figure 7.3 are not the same so −x4 6= (−x)4 .
1
(−x)4
x
−1
−x4
1
−1
Figure 7.3
Solutions for Section 7.2
EXERCISES
1. This is a power function.
2. This is not a power function. In a power function, the exponent is a constant, not a variable.
3. This is not a power function. It is the sum of two power functions.
4. This is a power function. We can rewrite it as y = 2x−3 .
5. This is a power function. We can rewrite it as y = 21 x3 .
6. This is a power function. We can rewrite it as y = 2x2 .
√
7. We have 3 p = 3p1/2 . The coefficient is 3 and the exponent is 1/2.
√
8. We have 2b = (2b)1/2 = 21/2 b1/2 . The coefficient is 21/2 and the exponent is 1/2.
4
4
= 1/4 = 4z −1/4 . The coefficient is 4 and the exponent is −1/4.
9. We have √
4
z
z
√
x
x1/2
10. We have √
=
= x(1/2)−(1/3) = x1/6 . The coefficient is 1 and the exponent is 1/6.
3
x1/3
x
1 x4
1 x4
1
1
x4
=
= x4−1 = x3 . The coefficient is 1/4 and the exponent is 3.
11. Since x > 0 have √ =
1/2
2
4 (x2 )
4 x
4
4
4 x
12. We have
1
√
5 x
r
3
=
53
1 −3/2
1
1
=
x
. The coefficient is 1/125 and the exponent is −3/2.
=
3
3/2
1/2
125
125x
(x )
81/3
8
2
=
= 2 = 2x−2 . The coefficient is 2 and the exponent is −2.
6
x
x
(x6 )1/3
√
14. We have (2 x)x2 = 2x1/2 x2 = 2x2+(1/2) = 2x5/2 . The coefficient is 2 and the exponent is 5/2.
r√
1/2 1/2
31/4 s1/2
3
31/4 s1/2
3
=
15. We have
s=
s
. The coefficient is 31/4 /2 and the exponent is 1/2.
=
4
4
2
41/2
13. We have
3
7.2 SOLUTIONS
16. We have
is 3/4.
p√
4t3 = (4t3 )1/2
1/2
= 41/2 t3/2
1/2
17. We have
= 2t3/2
1/2
219
= 21/2 t3/4 . The coefficient is 21/2 and the exponent
2
√ = (2/3)x−1/2 ,
3 x
so k = 2/3 and p = −1/2.
18. We have
1
= (1/3)x−2 ,
3x2
so k = 1/3 and p = −2.
19. We have
(7x3 )2 = 72 (x3 )2 = 49x6 .
We see that k = 49 and p = 6.
20. We have
r
√
49
49
7
√
=
= 5/2 = 7x−5/2 .
5
x5
x
x
Thus k = 7 and p = −5/2.
21. We have
(3x2 )3 = 33 (x2 )3 = 27x6 .
Thus k = 27 and p = 6.
22. We have
2
√
x
3
=
23
(x1/2 )3
=
8
= 8x−3/2 .
x3/2
Thus k = 8 and p = −3/2.
23. This represents proportionality to a power:
y=
1
1
= x−1 .
5x
5
Thus k = 1/5 and p = −1.
24. Since x2 and 3x2 are terms with the same power of x, we begin by combining like terms. We have
(x2 + 3x2 )2 = (4x2 )2 = 42 (x2 )2 = 16x4 .
We see that k = 16 and p = 4.
25. Since 3 and x2 are not like terms, we cannot combine them. It is not possible to write this expression in the form kxp .
26. We have
√
9x5 =
√
3
=
We see that k = 3 and p = 5/2.
27. We have
1
√
2 x
9·
√
x5 = 3(x5 )1/2 = 3x5/2 .
1
13
1
=
√
= x−3/2 .
8
8x3/2
23 ( x)3
We see that k = 1/8 and p = −3/2.
28. Since x2 and 4x are not like terms, we cannot combine them. It is not possible to write this expression in the form kxp .
29. We have
(−2x)2
3
= (−2x(−2x))3
= 4x2
3
= 4x2 · 4x2 · 4x2
= 64x6 ,
the power is p = 6, and the coefficient is k = 64.
220
Chapter Seven /SOLUTIONS
30.
√
3
x
x/8 = √
3
8
x1/3
=
2
1 1/3
= ·x ,
2
p
3
so k = 1/2, p = 1/3.
31. Writing this as
1
x3
= x3 ,
3
3
we have k = 1/3 and p = 3. If a > b, then a3 > b3 and k is positive. So f (a) > f (b).
32. Writing this as
we have k =
√
33.
√
√
√
1
5 · 7 = 5 · t−7 ,
t
< b−7 and k is positive. So f (a) < f (b).
5
t7
5, p = −7. If a > b, then a−7
=
r
so k =
√
34.
√
12
12
= √
r
r
√
1
= 12 · 1/2
r
√
= 12r −1/2 ,
12, p = −1/2. If a > b, then a−1/2 < b−1/2 and k is positive. So f (a) < f (b).
−3t2 −2t3
5
= −3t2 (−2)5 t3
= −3t2 (−32)t3·5
5
= 96t2+15
= 96t17 ,
so k = 96, p = 17. If a > b, then a17 > b17 and k is positive. So f (a) > f (b).
35. Writing this as
−5 (−2v)2
3
= −320v 6 ,
= −5 (−2)2 v 2
= −5(4)3 (v 2 )3
3
= −5 · 64 · v 6
we have k = −320, p = 6. If a > b, then a6 > b6 . But k is negative and so f (a) < f (b).
36. Writing this as
√
√
1
3
9
2 3 x + 3 x3 = 2x 3 + 3x 9
1
1
= 2x 3 + 3x 3
1
3
= 5x ,
we have k = 5, p = 1/3. If a > b, then a
1/3
>b
1/3
and k is positive. So f (a) > f (b).
PROBLEMS
37. This is a power function with exponent 1 so the graph is a line and graph (III) fits best.
38. Writing this as N = kA1/3 , we see that this is a power function with exponent 1/3, so graph (IV) fits best.
7.2 SOLUTIONS
221
39. Energy E is a power function of v with exponent 3, so graph (I) fits best.
40. Strength S is a power function of h with exponent 2, so graph (I) fits best.
41. Writing this as v = dt−1 . we see that v is a power function of t with exponent −1. Since the exponent is negative, graph
(II) fits best.
42. We see that S is a power function of B with exponent 2/3, so graph (IV) fits best.
43. Writing this as N = AL−2 , we see that N is a power function of L with exponent −2. Since the exponent is negative,
graph (II) fits best.
44. Since T = mB 1/4 , we see that T is a power function of B with exponent 1/4. Graph (IV) fits best.
45. We see that W is a power function of L with exponent 3, so graph (I) fits best.
46. Since s = (0.01h0.75 ) · w0.25 , we see that s is a power function of w with exponent 0.25. Graph (IV) fits best.
47. The judged loudness J is a power function of L with exponent 0.3 so graph (IV) fits best.
48. This is a power function with exponent 1 so the graph is a line and graph (III) fits best.
49. Negative. Since the expression does not equal zero, and since the square of any non-zero number is positive, we have
−(b − 1)2 = − (A positive number)
= A negative number.
√
50. Negative. The expression does not equal zero, so c 6= 0. Since −c is defined, −c must be positive, so c itself must be
negative. Because the square root of a positive number is positive, this means
√
c −c = (A negative number) · (A positive number)
= A negative number.
51. Positive. Since r 2 > 0 for any value of r, we have
1 + r2
2
− 1 = (1 + A positive number)2 − 1
= (A number larger than 1)2 − 1
= (A number larger than 1) − 1
= A positive number.
52. Negative. √
Note that neither v nor w equal 0, for otherwise the expression equals zero.
√ This means v is negative, for
otherwise −v is undefined. Since v is negative, w must also be negative, for otherwise vw is undefined. We know that
the square root of any positive number is positive, so we have
√
√
w −v + v vw = A negative number · A positive number + A negative number · A positive number
= A negative number + A negative number
= A negative number.
53. (a) The term in l and the term in w cannot be combined, so we cannot write this expression as a power function.
(b) Since l = 3w, we have
P = 2 · 3w + 2w = 8w.
So P is a power function of w, with k = 8, and p = 1.
54. (a) We have
π
A=l +
2
2
2
l
4
= l2 +
π
π 2
l = 1+
8
8
l2 .
So A is a power function of l, with k = 1 + π/8 and p = 2.
(b) The area is larger when l = 2.5 feet because (2.5)2 > (1.5)2 and the coefficient k is positive.
222
Chapter Seven /SOLUTIONS
55. Rewriting this as
s=
r
4
√ A=
3
r
2
4 √
√ A = 1/4 A1/2 ,
3
3
we see that the coefficient is 2/31/4 and the exponent is 1/2.
56. Remains unchanged, since
(ax)2
a2 x2
= 2 = x2 .
2
a
a
57. Increases, since (ax)2 + a2 = a2 (x2 + 1).
58. Decreases, since
(ax)2
a2 x2
x2
=
=
.
a3
a3
a
59. Increases.
60. Increases, since
1
= ax .
a−x
61. Decreases, since
x + a−1 = x +
1
.
a
62. Remains unchanged, since a0 x = x.
63. Remains unchanged, since
x+
64. Increases, since
1
= x + 1.
a0
a4
a
a4
= 3 3 = 3.
3
(ax)
a x
x
65. Decreases, since
(ax)1/3
a1/3 x1/3
x1/3
√
=
= 1/6 .
1/2
a
a
a
66. For investment A, P = $7000, r = 4%, and t = 5 years, so the value of the certificate will be
7000(1 + 0.04)5 = 8516.57.
For investment B, P = $6500, r = 6% and t = 7 so the value of the certificate will be
6500(1 + 0.06)7 = 9773.60.
Therefore the second investment is worth more.
67. For investment A, P = $10,000, r = 2%, and t = 10 years, so the value of the certificate will be
$10,000(1 + 0.02)10 = 12,189.94.
For investment B, P = $5000, r = 4%, and t = 10 years, so the value of the certificate will be
5000(1 + 0.04)10 = 740 = 7401.22.
Therefore the first investment is worth more.
7.3 SOLUTIONS
223
68. For investment A, P = $10,000, r = 3%, and t = 30 years, so the value of the certificate will be
10,000(1 + 0.03)30 = 24,272.62.
For investment B, P = $15,000, r = 4%, and t = 15 years, so the value of the certificate will be
15,000(1 + 0.04)15 = 27,014.15.
Therefore the second investment is worth more.
69. (a) We have x = 2% = 0.02, so
Population = 15(1 + 0.02)20 = 22.289,
so the town has 22,289 people in 20 years.
(b) We have x = 7% = 0.07, so
Population = 15(1 + 0.07)20 = 58.045,
so the town has 58,045 people in 20 years.
(c) We have x = −5% = −0.05, so
Population = 15(1 − 0.05)20 = 5.377,
so the town has 5,377 people in 20 years.
70. (a) Since r is expressed in thousands of miles, we express the other distances in the same units. The distance from the
center to the surface is 4 thousand miles, and the distance from the surface to the astronaut is h miles, or h/1000
thousand miles. So the total distance from the center to the astronaut is
h
r =4+
.
1000
(b) Since w = 2880/r 2 , we have
2880
w=
.
(4 + h/1000)2
Solutions for Section 7.3
EXERCISES
1. We raise both sides to the 1/3 power to obtain:
x = 501/3 = 3.684.
2. We first solve for x2 , then take the square root of both sides. Remember to include both the positive and negative roots:
2x2 = 8.6
x2 = 8.6/2 = 4.3
√
x = ± 4.3 = ±2.074.
There are two solutions: x = 2.074 and x = −2.074.
3. We raise both sides to the −2 power:
4 = x−1/2
4−2 = (x−1/2 )−2 = x1 = x
x = 4−2 = 1/(42 ) = 1/16.
4. We solve for w3 and then raise both sides to the 1/3 power:
4w3 = −7
w3 = −7/4 = −1.75
w = (−1.75)1/3 = −1.205.
224
Chapter Seven /SOLUTIONS
5. When we solve for z 2 , we obtain z 2 = −5. If we take the square root of both sides, we try to evaluate
real solutions. This equation has no solutions.
√
−5 which has no
6. We solve for b4 and then raise both sides to the 1/4 power. Since we are finding an even root, we need to remember to
include both the positive and negative roots.
2b4 = 92
b4 = 92/2 = 46
b = ±461/4 = ±2.604.
There are two solutions: b = 2.604 and b = −2.604.
√
7. We solve for a and then square both sides:
√
a=9
a = 92 = 81.
8. We solve for
√
3
x and then raise both sides to the 3rd power.
√
3 3 x = 15
√
3
x = 15/3 = 5
x = 53 = 125.
p
9.
y − 2 = 11
y − 2 = 121
y = 123.
p
10.
2y − 1 = 9
2y − 1 = 81
2y = 82
y = 41.
11.
√
3x − 2 + 1 = 10
√
3x − 2 = 9
3x − 2 = 81
3x = 83
83
.
x=
3
12. We solve for the power (x + 1)2 and then take the square root of both sides. Since we are taking an even root, we need to
remember to include both the positive and negative roots.
(x + 1)2 = 25
√
x + 1 = ± 25 = ±5.
We have x + 1 = 5 which gives x = 4 and also x + 1 = −5 which gives x = −6. There are two solutions:
x=4
and
x = −6.
7.3 SOLUTIONS
13. We solve for the power (3c − 2)3 , then take the cube root of both sides and then solve for c.
(3c − 2)3 = 150
3c − 2 = (150)1/3 = 5.313
3c = 7.313
c = 7.313/3 = 2.438.
14. We rewrite the equation with positive exponents. Then we take the cube root of both sides and solve it for x.
2x = 54x−2
54
2x = 2
x
2x3 = 54
x3 = 27
x = (27)1/3 = 3.
15. We solve for p5 then take the fifth root of both sides.
2p5 + 64 = 0
2p5 = −64
p5 = −32
p = (−32)1/5
p = −2.
16. First we put the powers of t on one side and the constants on the other. Then we take the fourth root of both sides.
4
1 3
t =
4
t
1 4
t =4
4
t4 = 16
t = ±(161/4 )
t = ±2.
17.
1
=0
L2
1
16 = 2
L
16L2 = 1
1
L2 =
16
1
L=± .
4
16 −
225
226
Chapter Seven /SOLUTIONS
18. We can solve this equation by squaring both sides.
p
r 2 + 144 = 13
r 2 + 144 = 169
r 2 = 25
r = ±5.
Because we squared the original
√ equation (which is not a reversible step) we need to check that the values we got r = 5
and r = −5 actually satisfy r 2 + 144 = 13. They do, so they are both solutions.
19. We can solve this equation by cubing both sides.
1
√
= −3
3
x
1
√
3
x
3
= (−3)3
1
= −27
x
1
x=− .
27
20. First we combine similar terms and solve for
√
x. We then square both sides.
√
√
2
4 x−2 x = x
3
√
2
2 x= x
3
√
1
x= x
3
1√
1=
x
3
√
3= x
x = 32 = 9.
21. We can solve this equation by squaring both sides.
12 =
q
z
5π
720π = z.
z
5π
144 =
Because we squared the original
q equation (which is not a reversible step) we need to check that the value we got, z =
z
720π, actually satisfies 12 =
. It does, so it is a solution.
5π
22. We divide both sides by k and then take the square root of both sides. Since we are taking an even root, we have to
remember to include both the positive and negative roots.
y = kx2
y
= x2
k
x=±
q
y
.
k
7.3 SOLUTIONS
227
23. We solve for r 2 and then take the square root of both sides. Since we are taking an even root, we have to remember to
include both the positive and negative roots.
1
A = πr 2
2
2A
2
r =
π
r=±
r
2A
.
π
If we knew that r had to be non-negative (for example, if r represented the radius of a circle), then we would only include
the positive root.
24. We solve for D3 and then raise both sides to the 1/3 power.
L = kB 2 D3
L
D3 =
kB 2
L 1/3
D=
.
kB 2
25. If x 6= 0, first we simplfy the right side of the equation then divide by y 2 in order to solve for x. We then take the square
root of both sides.
y 2 x2 = (3y 2 )2
y 2 x2 = 9y 4
x2 = 9y 2
x=±
p
9y 2 = ±3y.
26. We begin by squaring both sides of the equation in order to eliminate the radical.
w = 4π
q
w2 = 16π 2
x
t
x
t
tw2
= x.
16π 2
Because we squared the original equation
(which is not a reversible step) we need to check that the value we got,
q
x
2
2
. It does, so it is a solution.
tw /16π , actually satisfies w = 4π
t
27. (i), because the cube root of a positive is positive.
28. (i), because the fifth root of a positive is positive.
29. (iv), because positive numbers have two square roots.
30. (iii), because the square root of 0 is 0.
31. (vi), because no real number squared is negative.
32. (ii), because the cube root of a negative is negative.
33. (vi), because even powers of real numbers are not negative.
34. (ii), because an odd power of x is negative only if x is negative.
35. (i), because the cube root of x is positive only if x is positive.
36. (ii), because the cube root of x is negative only if x is negative.
37. (i), because the square root of x is positive only if x is positive.
38. (vi), because the square root of x is non-negative. Note that this equation is different from the equation x2 = 144, which
has two solutions, written 1441/2 and −(1441/2 ).
228
Chapter Seven /SOLUTIONS
39. (i), because only a positive number to an odd power is positive.
40. (iv), because both positives and negatives raised to even powers are positive.
41. (ii), because only negatives have negative odd powers.
42. (vi), because no even power is negative.
43. The second equation results from squaring both sides of the first equation. We solve the second equation
x + 4 = (x − 2)2
x + 4 = x2 − 4x + 4
5x − x2 = 0
x(5 − x) = 0
so x = 5 and x = 0 are solutions. Checking both of these solutions in the first equation gives
√
5+4 = 5−2
√
9=3
3=3
and
√
0+4 = 0−2
√
4 = −2
2 6= −2.
The extraneous solution, x = 0, was introduced by the squaring operation.
44. The second equation results from squaring both sides of the first equation. We solve the second equation
(t + 1)2 = 1
2
t + 2t + 1 = 1
t2 + 2t = 0
t(t + 2) = 0
so t = 0 and t = −2 are solutions. Checking both these solutions in the first equation gives
0+1 = 1
1=1
and
−2 + 1 = 1
−1 6= 1.
The extraneous solution, t = −2, was introduced by the squaring operation.
45. The second equation results from dividing both sides of the first equation by the variable r. To check whether the two
equations have the same solutions, we solve the first equation
r 2 + 3r = 7r
r 2 − 4r = 0
r(r − 4) = 0
so r = 4 and r = 0 are solutions. Checking both solutions in the second equation gives
4+3 = 7
7=7
7.3 SOLUTIONS
229
and
0+3 = 7
3 6= 7.
The two equations are not equivalent because they do not have the same solutions. The solution r = 0 was lost in
the operation of dividing by the variable r.
46. The second equation results from improperly taking the square root of both sides. To check whether the two equations
have the same solutions, we note that the first equation has x = ±2 as solutions but the second equation has only x = 2
as a solution. The two equations are not equivalent and x = −2 was lost in the operation. The correct result of taking the
square root of both sides of the first equation would have yielded 2x = ±4 and preserved both solutions.
47. The second equation results from multiplying both sides of the first equation by the variable p. Solving the second equation
we find,
3p − 1 = p − 1
2p = 0
p = 0.
However, p = 0 is not a solution of the first equation because the denominators are zero for p = 0, thus p = 0 is an
extraneous solution.
48. The second equation results from multiplying both sides of the first equation by x + 1. Solving the second equation, we
find,
2x = x + 1 − 2
2x = x − 1
x = −1.
However, x = −1 is not a solution of the first equation because the denominators are zero for x = −1, thus x = −1 is
an extraneous solution.
PROBLEMS
49. Both solutions must be positive, for if x = 0, the equation clearly does not work, and if x < 0, then
√
x 8 − x = A negative number × A positive number
|
{z
x
}
= A negative number,
|
making the value of the left-hand side less than 5.
50. Since the volume, V is 3π, and since V = 32 πr 2 , we have
2 2
πr
3
3
3 2
· 3π = · πr 2
2
2 3
9
π = πr 2
2
1
1 9
· π = · πr 2
π 2
π
9
= r2
2
r
√
9
±
= r2
2
V = 3π =
{z
√
8−x
}
230
Chapter Seven /SOLUTIONS
3
± √ = r.
2
√
Since a cone’s radius cannot be negative, r = 3/( 2) = 2.121.
51. Since the volume, V , is 27 cm3 and V = s3 , we have
V = 27 = s3
√
√
3
3
27 = s3
3 = s.
The side length is 3 cm for a cube of volume 27 cm3 .
52. The solution to the equation represents the price charged when the electricity consumption is 2 gigawatt-hours per year.
To find the solution, we solve for p:
2=
0.15
p3/2
2p3/2 = 0.15
0.15
p3/2 =
2
0.15 2/3
p=
= 0.18 dollars per kilowatt-hour.
2
Check that 0.18 is a solution to the original equation.
53. (a) We have
S = 1095M 2/3 .
We substitute S = 21,000 and solve for M :
21,000 = 1095M 2/3
21,000
= M 2/3
1095
21,000 3/2
= 83.986 kg.
M =
1095
The body mass is about 84 kg.
(b) The solution represents the body mass of a human with surface area 30,000 cm2 .
(c) We have
S = 1095M 2/3
1
S = M 2/3 ,
1095
so
M=
The formula is
1
S
1095
3/2
=
1
1095
3/2
· S 3/2 = 0.0000276S 3/2 .
M = 0.0000276S 3/2 .
54. Because r > 0, we will ignore any negative solutions for r.
(a) We have
r=
(b) We have
r=
1
C.
2π
r
1
A.
π
7.3 SOLUTIONS
(c) We have
r=
(d) We have
3
V
4π
r=
(e) We have
r=
r
r
1/3
231
.
V
.
πh
3
V.
πh
1
4 3
πr = πr 2 h.
3
3
(b) We solve the equation in part (a) by canceling (1/3)πr 2 from both sides, giving h = 4r. Thus, the cone should be
twice as high as it is wide.
55. (a) For the sphere and the cone to have the same volume, we want
56. (a) When r = 0, we have A = P . This makes sense, because if the interest rate is 0%, there is no interest being added,
and the amount remains the same as the initial amount, P .
(b) If the interest rate is 5%, then A = P (1.05)2 = P (1.1025). If the interest rate is 6%, then A = P (1.06)2 =
P (1.1236). So if the interest rate is between 5% and 6%, then the percentage growth in the amount after 2 years will
be between 10.25% and 12.36%.
p
p
(c) A = P (1+r)2 , so A/P = (1+r)2 . Taking the square root of both sides, we get A/P = 1+r, or r = A/P −1.
This form of the equation might be useful if you have a goal for the amount of money that you would like to have
saved at the end of 2 years, and you want to find out at what interest rate you would need to invest your principal in
order to achieve that goal.
(d) If the principal amount
A will be at least 1.25P . Using this value of A
p increases by 25% by the
p end of 2 years, then
√
in the equation r = A/P − 1, we get r = 1.25P/P − 1 = 1.25 − 1 = 0.118. So r must be 11.8%.
57. After t years, the balance B is given by
r t
,
100
where P is the initial deposit. We know that B = 2P when t = 10, so
B =P 1+
2P = P 1 +
Dividing by P gives
2= 1+
Solving for r, we take roots
1+
r
100
r
100
10
10
.
.
r
= 21/10
100
r
= 21/10 − 1
100
r = 100(21/10 − 1) = 7.177.
Thus the interest rate is 7.177%.
√
√
58. For x to be meaningful, we need x ≥ 0, in which case x + 5 x ≥ 0, which cannot be negative.
59. (a) If t = 0 then t3 = 0, so A = 0.
(b) If t > 0 then t3 > 0, so A > 0.
(c) If t < 0 then t3 < 0, so A < 0.
60. (a) If t = 0 then t4 = 0, so A = 0.
(b) If t > 0 then t4 > 0, so A > 0.
(c) If t < 0 then t4 > 0, so A > 0.
232
Chapter Seven /SOLUTIONS
61. (a) If t = 0 then (−t)3 = 0, so A = 0.
(b) If t > 0 then (−t)3 = (−1)3 t3 = −t3 < 0, so A < 0.
(c) If t < 0 then (−t)3 = (−1)3 t3 = −t3 > 0, so A > 0.
62. (a) If t = 0 then (−t)4 = 04 = 0, so A = 0.
(b) If t > 0 then (−t)4 = (−1)4 t4 = t4 > 0, so A > 0.
(c) If t < 0 then (−t)4 = (−1)4 t4 = t4 > 0, so A > 0.
63. (a) If t = 0 then t3 = 0, so A2 = 0, that is A = 0.
(b) If t > 0 then t3 > 0, so A2 > 0, which is satisfied for any A 6= 0.
(c) If t < 0 then t3 < 0, but A2 > 0, which is not satisfied for any A.
64. (a) If t = 0 then t4 = 0, so A2 = 0, that is A = 0.
(b) If t > 0 then t4 > 0, so A2 > 0, which is satisfied for any A 6= 0.
(c) If t < 0 then t4 > 0, so A2 > 0, which is satisfied for any A 6= 0.
65. (a) If t = 0 then t4 = 0, so −A2 = 0, that is A = 0.
(b) If t > 0 then t4 > 0, but −A2 < 0, which is not satisfied for any A.
(c) If t < 0 then t4 > 0, so −A2 < 0, which is not satisfied for any A.
66. (a) If t = 0 then t3 = 0, so −A2 = 0, that is A = 0.
(b) If t > 0 then t3 > 0, but −A2 < 0, which is not satisfied for any A.
(c) If t < 0 then t3 < 0, so −A2 < 0, which is satisfied for any A 6= 0.
67. (a) If t = 0 then 1 = A, so A = 1.
(b) If t > 0 then t3 + 1 > 1, so A > 1.
(c) If t < 0 then t3 + 1 < 1, so A < 1.
68. The equation At2 + 1 = 0 is the same as At2 = −1.
(a) If t = 0 then 0 = −1, which means that no value of A will work.
(b) If t > 0 then t2 > 0, so we need A < 0 to make At2 < 0.
(c) If t < 0 then t2 > 0, so we need A < 0 to make At2 < 0.
69. (a) If t = 0 then At2 = 0 for any A.
(b) If t > 0 then t2 > 0, so At2 = 0 only if A = 0.
(c) If t < 0 then t2 > 0, so At2 = 0 only if A = 0.
70. The equation A2 t2 + 1 = 0 is the same as A2 t2 = −1. The left-hand side is never negative, while the right-hand side is
always negative. Thus, no values of A can make this true.
(a) No A.
(b) No A.
(c) No A.
71. (a) If x = 1 than A = 4.
(b) If x > 1 then A > 4.
(c) There is no solution for A < 0.
72. (a) If x = 1 then A = 4.
(b) If x > 1 then A > 4.
(c) This equation has a solution for all values of A. So no A.
73. (a) If x = 1 then A = −4.
(b) If x > 1 then A < −4.
(c) This equation has no solution for A > 0.
74. (a) If x = 1 then A = 4.
(b) If x > 1 then 0 < A < 4.
(c) This equation has no solution for A ≤ 0.
75. (a) We see that the graphs cross in the first quadrant. We can find the point of intersection more accurately by zooming
in on the graph. We see that the point of intersection occurs at approximately x = 9.6. See Figure 7.4.
7.3 SOLUTIONS
233
y
1200
x = 9.6
1100
x
9
10
Figure 7.4
(b) Using algebra, we have
1.3x3 = 120x
120
x2 =
1.3
x=
r
120
= 9.608.
1.3
We used only the positive square root since we wanted the point of intersection in the first quadrant.
76. (a) We see that the graphs cross in the first quadrant. We can find the point of intersection more accurately by zooming
in on the graph. We see that the point of intersection occurs at approximately x = 17.1. See Figure 7.5.
y
300000
x = 17.1
290000
x
16
17
18
Figure 7.5
(b) Using algebra, we have
0.2x5 = 1000x2
1000
= 5000
x3 =
0.2
x = (5000)1/3 = 17.100.
234
Chapter Seven /SOLUTIONS
Solutions for Section 7.4
EXERCISES
1. We have y = kx5 . We substitute y = 744 and x = 2 and solve for k:
y = kx5
744 = k(25 )
744
744
k= 5 =
= 23.25.
2
32
The formula is
y = 23.25x5 .
2. We have y = kx3 , with k = −0.35. The formula is
y = −0.35x3 .
3. We have y = kx2 . We substitute y = 1000 and x = 5 and solve for k:
y = kx2
1000 = k(52 )
1000
1000
= 40.
k= 2 =
5
25
The formula is
y = 40x2 .
4. We have y = kx4 . We substitute y = 10.125 and x = 3 and solve for k:
y = kx4
10.125 = k(34 )
10.125
10.125
k=
= 0.125.
=
34
81
The formula is
y = 0.125x4 .
√
5. We have s = k t. We substitute s = 100 and t = 50 and solve for k:
√
s=k t
√
100 = k 50
100
k = √ = 14.142.
50
The formula is
√
s = 14.142 t.
6. We have A = k/B 3 . We substitute A = 20.5 and B = −4 and solve for k:
k
B3
k
20.5 =
(−4)3
A=
k = 20.5 · (−4)3 = −1312.
The function is
A=−
1312
.
B3
7.4 SOLUTIONS
7. Substituting into the general formula c = kd2 , we have 50 = k(5)2 or k = 50/25 = 2. So the formula for c is
c = 2d2 .
When d = 7, we get c = 2(7)2 = 98.
8. Substituting into the general formula c = k/d2 , we have 50 = k/52 or k = 1250. So the formula for c is
c=
1250
.
d2
When d = 7, we get c = 1250/72 = 25.510.
9. For some constant k, we have S = kh2 .
10. We know that E is proportional to v 3 , so E = kv 3 , for some constant k.
√
11. We have r = k A = kA1/2 .
12. For some constant k, we have K = kv 2 .
13. When x is doubled, we have
new value of y = k(2x)3 = k · 23 x3 = 8(kx3 ).
So y is multiplied by a factor of 8.
14. When x is doubled, we have
new value of y =
k
k
1 k
=
= · 3.
(2x)3
8x3
8 x
So y is multiplied by a factor of 1/8.
15. Dividing both sides by x, we get
y=
k
.
x
When x is doubled, we have
new value of y =
k
1 k
= · .
2x
2 x
So y is multiplied by a factor of 1/2.
16. Multiplying both sides by x4 , we get
When x is doubled, we have
y = kx4 .
new value of y = k(2x)4 = k · 24 x4 = 16(kx4 ).
So y is multiplied by a factor of 16.
PROBLEMS
17. (a) We have T = kR2 D4 .
(b) We substitute R = 300D and simplify:
T = kR2 D4 = k(300D)2 D4 = 90,000kD6 .
So T is a power function√of D.
(c) We substitute D = 0.25 R and simplify:
√
T = kR2 D4 = kR2 (0.25 R)4 = (0.25)4 kR2 R2 = 0.0039kR4 .
So T is a power function of R.
235
236
Chapter Seven /SOLUTIONS
18. (a) Since the rate R varies directly with the fourth power of the radius r, we have the formula
R = kr 4 ,
where k is a constant.
(b) Given R = 400 for r = 3, we can determine the constant k.
400 = k(3)4
400 = k(81)
400
k=
= 4.938.
81
So the formula is
R = 4.938r 4 .
(c) Evaluating the formula at r = 5 yields
R = 4.938(5)4 = 3086.42
cm 3
.
sec
19. (a) T is proportional to the fourth root of B, and so
√
4
T = k B = kB 1/4 .
(b) 148 = k · (5230)1/4 and so k = 148/(5230)1/4 = 17.4.
(c) Since T = 17.4B 1/4 , for a human with B = 70 we have
T = 17.4(70)1/4 = 50.3 seconds.
It takes about 50 seconds for all the blood in the body to circulate and return to the heart.
20. (a) We have R = kV 2 , where k is the constant of proportionality. Thus, R = 0.1639 (210)2 = 7228 feet = 1.37 miles.
(b) If R is now measured in meters, R = 7228 ft = (7228/3.28) meters = 2204 meters, and V = 210 miles/hour =
210(1609 meters)/3600 sec = 93.86 meters/sec. So the new constant of proportionality is 2204/93.862 = 0.25.
21. Since N is inversely proportional to the square of L, we have
N=
k
.
L2
As L increases, N decreases, so there are more species at small lengths.
22. We need to solve y = kxp for p and k. We know that y = 2 when x = 1. Since we also have y = k · 1p = k when x = 1,
we have k = 2. To solve for p, use the fact that y = 8 when x = 2 and also y = 2 · 2p when x = 2, so
2 · 2p = 8,
giving 2p = 4, so p = 2. Thus, y = 2x2 . We must check that all the points in the table satisfy this equation. They do.
23. We need to solve y = kxp for p and k. To solve for p, take the ratio of any two values of y, say the values corresponding
to x = 3 and x = 2, namely y = 36 and y = 16:
9
36
= .
16
4
Since y = k · 3p when x = 3 and y = k · 2p when x = 2, we have
k · 3p
3p
= p =
k · 2p
2
p
3
2
=
9
.
4
Since ( 23 )p = 49 , we know p = 2. Thus, y = kx2 . To solve for k, use any point from the table. Since y = 16 when x = 2,
we have k · 22 = 4k = 16, so k = 4. Thus, y = 4 · x2 . We must check that all the points in the table satisfy this equation.
They do.
7.4 SOLUTIONS
237
24. Solve for y by taking the ratio of (say) the values of y when x = 1 and when x = 2:
−16
= 16.
−1
We know y = k · 2p when x = 2 and y = k · 1p when x = 1. Thus,
k · 2p
2p
=
=
k · 1p
1p
p
2
1
= 16.
Thus p = 4. To solve for k, note that y = k · 24 = 16k when x = 2. Thus, 16k = −16. Thus, k = −1. This gives
y = −x4 . We must check that all the points in the table satisfy this equation. They do.
25. Solve for y by taking the ratio of, say, the values of y when x = 2 and x = 1
−8/5
= 8.
−1/5
We know y = k · 2p when x = 2 and y = k · 1p when x = 1. Thus,
k · 2p
= 2p = 8.
k · 1p
Then p = 3. To solve for k, note that y = k · 2p = k · 23 = 8k when x = 2. Thus we have 8k = − 58 . Then k = −1/5,
which gives y = −(1/5)x3 . We must check that all the points in the table satisfy this equation. They do.
26. (a) We have
f (2r)
=
f (r)
=
4
π(2r)3
3
4
πr 3
3
4
π 8r 3
3
·
4
π r3
3
= 8.
(b) We have
f (r)
=
f ( 12 r)
=
=
4
πr 3
3
4
π( 12 r)3
3
4
π r3
3
·
4
π 18 r 3
3
1
1
8
= 8.
(c) Both expressions equal 8 which means that in both cases, the larger sphere has eight times the volume of the smaller.
• The first result tells us that the volume of a sphere of radius 2r is eight times the volume of a sphere of radius r.
• The second result tells us that the volume of a sphere of radius r is eight times the volume of a sphere of radius
1
r.
2
• In both cases, the result shows that a sphere whose radius is twice that of a smaller sphere has eight times its
volume.
We see that, algebraically, it does not matter if we say the larger sphere has a radius twice the size of the smaller, or
instead the smaller sphere has a radius half the size of the larger.
27. (a) We have
g
r
10
g(r)
r −2
10
=
kmr −2
1
−2
km 10−2 · r
·
=
km
r −2
1
= 1 = 100.
km
100
238
Chapter Seven /SOLUTIONS
(b) We have
g(r)
kmr −2
=
g(10r)
km(10r)−2
r −2
km
· −2 −2
km 10 r
1
= 1 = 100.
=
100
(c) Both expressions equal 100 which means that in both cases, the force on the closer object is 100 times as great as the
force on the more distant object.
• The first result tells us that the force on an object at a distance of r/10 is 100 times the force at a distance of r.
• The second result tells us that the force on an object at a distance of r is 100 times the force at a distance of 10r.
• In both cases, the result shows that the force on an object is 100 times the force an an object of identical mass
that is ten times farther away.
This is telling us that, algebraically, it does not matter if we say the more distant object is at a distance ten times as
great, or instead the closer object is at a distance one-tenth as great.
28. When the side is x, the area is x2 .
(a) When the side is 2x,
Area = (2x)2 = 22 x2 = 4x2 .
So area is multiplied by a factor of 4.
(b) When the side is 3x,
Area = (3x)2 = 32 x2 = 9x2 .
So area is multiplied by a factor of 9.
(c) When the side is 21 x,
Area =
1 2
2
x
=
1 2
2
x2 =
1 2
x .
4
So area is multiplied by a factor of 1/4.
(d) When the side is 0.1x,
Area = (0.1x)2 = (0.1)2 x2 = 0.01x2 .
So area is multiplied by a factor of 0.01.
29. When the side is x, the volume is x3 .
(a) When the side is 2x,
Volume = (2x)3 = 23 x3 = 8x3 .
So volume is multiplied by a factor of 8.
(b) When the side is 3x,
Volume = (3x)3 = 33 x3 = 27x3 .
So volume is multiplied by a factor of 27.
(c) When the side is 21 x,
Volume =
1 3
2
x
=
1 3
2
x3 =
1 3
x .
8
So volume is multiplied by a factor of 1/8.
(d) When the side is 0.1x,
Volume = (0.1x)3 = (0.1)3 x3 = 0.001x3 .
So volume is multiplied by a factor of 0.001.
30. The area of a circle of radius r is πr 2 . If r is halved, then we get a new circle of radius (1/2)r, which has area
π
So the area is multiplied by 1/4.
1
r
2
2
=π
2
1
2
r2 =
1 2
πr .
4
7.4 SOLUTIONS
239
31. If the cube has side length s, then one of its faces has area s2 . Since there are six faces, the entire cube has surface area
6s2 . If s is increased by 10%, then the side length is multiplied by 1.1, so the new side length is (1.1)s, and the new
surface area is 6(1.1 · s)2 = 6(1.1)2 s2 = (1.1)2 (6s2 ) = 1.21 · 6s2 . So the surface area is increased by 21%.
32. If the cube has side length s, then it has volume s3 . If s is increased by 10%, then the side length is multiplied by 1.1, so
the new side length is (1.1)s, and the volume is
New volume = (1.1 · s)3 = (1.1)3 s3 = 1.331 · s3 = 1.331 Old volume.
So the volume is increased by 33.1%.
33. (a) For both birds and mammals the life span increases with body size. Thus the larger mammal will have the longer life
span. Because the graph of the bird life span is above that of the mammal, the bird has the greater life span, for a
fixed body size.
(b) A body size of 0 would mean no life span! From this and the graph we can construct the table and the corresponding
graph. This is very close to a line with slope 57/25 = 2.28. Thus, on average a bird will live over twice as long as a
mammal of the same size.
Table 7.1
Body size
0
10
20
30
40
Mammal life span
0
19
22
23
25
Bird life span
0
44
50
54
57
bird lifespan
60
40
20
10
20
30
mammal lifespan
Figure 7.6
(c)
(i) From LM = 11.8W 0.20 we find W =
0.19/0.20
L
M
11.8
1/0.20
which, when substituted in LB = 28.3W 0.19 , gives
LM
. This is the function that is plotted in part (b). Also, 0.19/0.20 = 0.95 is very
11.8
28.3
close to 1, so the graph looks like the line LB =
LM = 2.398LM .
11.8
28.3 100
=
(ii) To have LM = LB we would need 11.8W 0.20 = 28.3W 0.19 , or 11.8W 0.01 = 28.3, so W =
11.8
37
5
24
9.78 × 10 kg. (The largest animal is the blue whale which is about 10 kg. The earth weighs about 6 × 10
0.20
kg.) If there were an bird or mammal of this size, its life span would be 11.8 9.78 × 1037
= 4.6768 × 108
years. This is unrealistic.
LB = 28.3
34. If z is proportional to a power of y we have z = k1 y n , where k1 is a constant. If y is proportional to a power of x we have
y = k2 xm , where k2 is a constant. So z = k1 y n = k1 (k2 xm )n = (k1 k2n )xnm , so z is proportional to a power of x.
35. If z is proportional to a power of x we have z = k1 xn , where k1 is a constant. If y is proportional to the same power of
x we have y = k2 xn , where k2 is a constant. So z + y = k1 xn + k2 xn = (k1 + k2 )xn , so z + y is proportional to a
power x.
240
Chapter Seven /SOLUTIONS
36. If z is proportional to a power of x we have z = k1 xn , where k1 is a constant. If y is proportional to a power of x we
have y = k2 xm , where k2 is a constant. So zy = k1 xn · k2 xm = (k1 · k2 )xn+m , so zy is proportional to a power of x.
37. If z is proportional to a power of x we have z = k1 xn , where k1 is a constant. If y is proportional to a different power
of x we have y = k2 xm , where k2 is a constant and n 6= m. So z + y = k1 xn + k2 xm . Since n 6= m these two terms
cannot be combined to be proportional to a power of x, so z + y is not proportional to a power of x.
Solutions for Chapter 7 Review
EXERCISES
1. We have y = 3x−2 ; k = 3, p = −2.
2. We have y = 5x1/2 ; k = 5, p = 1/2.
3. We have y = 83 x−1 ; k = 38 , p = −1.
4. Not a power function.
5. We have y = 25 x−1/2 ; k = 52 , p = −1/2.
6. We have y = 9x10 ; k = 9, p = 10.
7. We have y = 0.2x2 ; k = 0.2, p = 2.
8. Not a power function.
9. We have y = 53 · x3 = 125x3 ; k = 125, p = 3.
10. We have y = 8x−1 ; k = 8, p = −1.
11. We have y = (1/5)x; k = 1/5, p = 1.
12. Not a power function because of the +4.
−1
−1 −1
13. Rewriting
as
w we see that w is the base, the exponent is −1 and −1/7 is the coefficient.
7w
7
14. When we raise (−2t) to the third power we obtain −8t3 . Therefore the base is t, the exponent is 3 and and the coefficient
is −8.
15. We have 125v 5 /25v 3 = 5v 2 , so the base is v, the exponent is 2 and the coefficient is 5.
4
16. The base is r, the exponent is 3, and the coefficient is π.
3
17. We know that (4x3 )(3x−2 ) = 12x. Therefore, the base is x, the exponent is 1, and the coefficient is 12.
18. The base is z, the exponent is 4, and the coefficient is −1.
3
19. We have 2 = 3x−2 . The base is x, the exponent is −2, and the coefficient is 3.
x
8
8x6
20. We have
=
= −4x6 . The base is x, the exponent is 6, and the coefficient is −4.
−2/x6
−2
21. We have 3(−4r)2 = 3(16r 2 ) = 48r 2 . The base is r, the exponent is 2, and the coefficient is 48.
5
2
1
1
= 5 = t−5 . The base is t, the exponent is −5, and the coefficient is 1/4.
10t5
4t
4
23. We have (πa)(πa) = π 2 a2 . The base is a, the exponent is 2, and the coefficient is π 2 .
22. We have
24. We have πa + πa = 2πa. The base is a, the exponent is 1, and the coefficient is 2π.
25. We multiply both sides by x3 , then solve for x3 and then raise both sides to the 1/3rd power:
50
= 2.8
x3
50 = 2.8x3
SOLUTIONS to Review Problems for Chapter Seven
50
2.8
1/3
50
x=
= 2.614.
2.8
x3 =
26. We have:
5
x2
5x3
x3
x2
x
8
x3
= 8x2
8
=
5
= 1.6.
=
27. We have:
12
√ =3
x
√
12 = 3 x
√
12
x=
=4
3
x = 42 = 16.
28. We have:
√
5
1
=
4
x−3
5√
1=
x−3
4
√
4
= x−3
5
x − 3 = (0.8)2
x = 3 + (0.8)2 = 3.64.
29. We have:
100
=4
(x − 2)2
100 = 4(x − 2)2
(x − 2)2 = 25
√
x − 2 = ± 25 = ±5.
This gives two solutions:
and
x − 2 = 5 so
x − 2 = −5 so
x = 7,
x = −3.
241
242
Chapter Seven /SOLUTIONS
30. We have:
A
=C
Bxn
A = BCxn
A
xn =
BC
A 1/n
.
x=
BC
31. (a) We have:
√
2π L
=R
C2
√
2π L = R · C 2
√
RC 2
L=
2π
L=
L=
R2 C 4
.
4π 2
RC 2
2π
2
(b) We have:
√
2π L
=R
C2
√
2π L = R · C 2
√
2π L
C2 =
R
C=±
r
√
2π L
.
R
32. Since we take an odd root to solve the equation, the equation has only one solution.
33. Since we take an even root to solve the equation, the equation has two solutions.
34. Since we square both sides to solve the equation, the equation has only one solution.
35. Since no real number when raised to an even power gives a negative number, this equation has no solutions.
36. Since we raise both sides of the equation to a power of 5, this equation has only one solution.
37. Since we take an odd root to solve the equation, this equation has only one solution.
PROBLEMS
38. (a) We have E = kLh2 .
(b) Since E = 4 million foot-pounds = 4,000,000 when L = 600 and h = 30, we have
4,000,000 = k · 600 · (302 )
4,000,000
k=
= 7.407.
600 · (302 )
We have
E = 7.407Lh2 .
SOLUTIONS to Review Problems for Chapter Seven
To find the units, we solve the equation E = kLh2 for k to obtain
k=
E
.
Lh2
The units of E are foot-pounds which are feet × pounds, so we have
Units of E
Units of L · (Units of h)2
foot-pounds
=
feet · feet2
feet × pounds
=
feet3
pounds
.
=
feet2
Units of k =
The units of k are pounds per square foot.
(c) We substitute h = (1/4)L = 0.25L and simplify:
E = 7.407Lh2 = 7.407L(0.25L)2 = 7.407(0.25)2 · L · L2 = 0.463L3 .
(d) We substitute L = 5h and simplify:
E = 7.407(5h)h2 = 37.037h3 .
39. We know R is proportional to the fourth power of r, with constant of proportionality 4.94 so we have
R = 4.94r 4 .
(a) We substitute R = 500 and solve for r:
500 = 4.94r 4
500
= r4
4.94
500 1/4
= 3.172 cm.
r=
4.94
The radius of the pipe is about 3.172 cm.
(b) We have
R = 4.94r 4
1
R = r4
4.94
r=
1
R
4.94
1/4
=
1
4.94
1/4
· R1/4 = 0.671R1/4 .
We have r = 0.671R1/4 .
(c) Since r =Constant ×R1/4 , we see that r is proportional to the 1/4th power of R.
40. We have
E = 7.4Lh2 .
(a) We substitute L = 50 and E = 40,000 and solve for h:
40,000 = 7.4(50)h2
40,000
= h2
7.4(50)
h=
r
40,000
= 10.398 ft.
7.4(50)
243
244
Chapter Seven /SOLUTIONS
(b) We substitute L = 20 and solve for h:
E = 7.4(20)h2
1
E
h2 =
7.4(20)
h=
r
1
E=
7.4(20)
r
√
1
· E = 0.0822E 1/2 .
7.4(20)
We have h = 0.0822E 1/2 . The coefficient is 0.0822 and the power is 1/2.
41. We have
P =
k
,
R3
for some constant k. We solve for R:
P · R3 = k
k
R3 =
P
R=
1/3
k
P
=
k1/3
Constant
=
.
P 1/3
P 1/3
The quantity R is inversely proportional to the cube root of P .
42. (a) We have T = kR2 D4 for some constant k.
(b) We solve for R:
T = kR2 D4
T
R2 =
kD4
R=
r
T 1/2
T
= 1/2 2 = (1/k1/2 )T 1/2 D−2 .
4
kD
k D
We have R = CT 1/2 D−2 so n = 1/2 and m = −2.
(c) We solve for D:
T = kR2 D4
T
D4 =
kR2
T 1/4
T 1/4
= 1/4 1/2 = (1/k1/4 )T 1/4 R−1/2 .
D=
2
kR
k R
We have D = CT 1/4 R−1/2 so n = 1/4 and m = −1/2.
43. (i), because for a positive power of x to be greater than 1, x must be greater than 1.
44. (i), because for a positive power of x to be greater than 1, x must be greater than 1.
45. (iii), because for a positive power of x to be less than 1, x must be less than 1.
46. (iii), because for a positive power of x to be less than 1, x must be less than 1.
47. (i), because for a positive power of x to be greater than 1, x must be greater than 1.
48. (iii), because for a positive power of x to be less than 1, x must be less than 1.
49. (i), because for a positive power of x to be greater than 1, x must be greater than 1.
50. (ii), because for a non-zero power of x to equal 1, x must equal 1.
51. (iii), because for a negative power of x to be greater than 1, x must be less than 1.
52. (i), because for a negative power of x to be less than 1, x must be greater than 1.
53. (i), because for a negative power of x to be less than 1, x must be greater than 1.
SOLUTIONS to Review Problems for Chapter Seven
245
54. (iv), because no even power is negative.
55. (iii), because for a negative power of x to be greater than 1, x must be less than 1.
56. (ii), because for a non-zero power of x to equal 1, x must equal 1.
57. Taking the square root of both sides of the equation, we get
√
2x = ± 16 = ±4
x = ±2.
58. Dividing both sides of the equation by 4, we get
x2 = 4
√
x = ± 4 = ±2.
59. Dividing both sides of the equation by 16, we get
x2
= 1.
4
Multiplying both sides by 4, we get
x2 = 4
√
x = ± 4 = ±2.
√
60. If we take the cube root of both sides, we cannot solve the equation because 3 x3 + 8 cannot be further simplified. Step (b)
is the appropriate step to solve the equation. After subtracting 8 from both sides, we can take the cube root of both sides
of the equation to get the solution.
61. Equations (c) and (d) have the same solutions as the given equation. Equation (c) is obtained from the given equation by
taking the square root of both sides. Equation (d) is obtained from the given equation by dividing both sides by 9. These
operations do not change the equality of the two sides and so the transformed equations have the same solutions as the
original given equation. Equation (a) does not have the same solutions as the given equation because the negative root is
not included. There is no way to obtain equation (b) from the given equation by carrying out the same operation on both
sides of the equation.
62. (a) See Figure 7.7.
(b) If s = 1, then the relation between d and w is given by w = kd. The graph of this equation looks like a straight line
through the origin, but the points on the plot do not lie on a straight line through the origin. So it does not support the
hypothesis.
(c) If s = 2 or 3, then we have w = kd2 or w = kd3 . Both of these look like graphs that go through the origin and then
curve upward. The points on the data plot also look like they lie on a curve that goes through the origin and curves
upward, so it is possible that s = 2 or s = 3.
(d)
Table 7.2
Diameter, d (cm)
5.6
6.5
11.8
16.7
23.4
Weight, w (kg)
5.636
7.364
30.696
76.730
169.290
w/d2
0.180
0.174
0.220
0.275
0.309
Overall, the ratios are increasing. If s = 2 were the correct value, the ratios should be all approximations of the
constant k, so they should all stay roughly the same.
(e)
Table 7.3
Diameter, d (cm)
5.6
6.5
11.8
16.7
23.4
Weight, w (kg)
5.636
7.364
30.696
76.730
169.290
w/d3
0.032
0.027
0.019
0.016
0.013
246
Chapter Seven /SOLUTIONS
This time the ratios are decreasing. In part (d) they were increasing overall. Somewhere in between we should
be able to find a value where they are roughly constant. So the correct value of s is somewhere between s = 2 and
s = 3.
(f) We calculate the ratios with s = 2.5.
Table 7.4
Diameter, d (cm)
5.6
6.5
11.8
16.7
23.4
Weight, w (kg)
5.636
7.364
30.696
76.730
169.290
w/d2.5
0.076
0.068
0.064
0.067
0.064
There seems to be still a slight downward trend in the ratios, so we try again with 2.4.
Table 7.5
Diameter, d (cm)
5.6
6.5
11.8
16.7
23.4
Weight, w (kg)
5.636
7.364
30.696
76.730
169.290
w/d2.4
0.090
0.082
0.082
0.089
0.088
Here the ratios show no clear upward or downward trend, so we estimate s = 2.4 and we take the average of the
ratios as our estimate for k, which is about 0.086. Thus the power function w = 0.086d2.4 is a good fit for the data,
as shown in Figure 7.8.
w (kg)
w (kg)
200
200
175
175
150
150
125
125
100
100
75
75
50
50
25
25
0
5
10
15
20
25
0
d (cm)
30
5
10
Figure 7.7
15
20
25
30
d (cm)
Figure 7.8
63. (a) Kepler’s Law states that the square of the period, P , is proportional to the cube of the distance, d. Thus, we have
P 2 = kd3 .
Solving for P gives
where we let k1 =
√
P =
√
kd3 =
√
kd3/2 = k1 d3/2 ,
k. For the earth, P = 365 and d = 93,000,000. Thus,
365 = k1 (93,000,000)3/2 ,
SOLUTIONS to Review Problems for Chapter Seven
so
k1 =
365
.
(93,000,000)3/2
This gives
P =
247
365
d3/2
· d3/2 = 365
= 365
3/2
(93,000,000)
(93,000,000)3/2
d
93,000,000
3/2
.
(b) For Mars, d = 142,000,000, so we have
P = 365
142,000,000
93,000,000
3/2
which gives
P = 689 earth days.
64. (a) Since the graph shows values of V that get larger as the values of n get larger, the exponent β must be positive. Since
it bends towards the horizontal axis, we have β < 1. So 0 < β < 1.
(b) The value of V corresponding to n = 1,000,000 is about V = 10,000. Since n0.5 = 1,000,0000.5 = 1000, we have
V = 10 · 1000 = 10n0.5 , so K = 10.
Solutions for Solving Drill
1. We raise both sides to the 1/3 power:
x3 = 10
x = 101/3 = 2.154.
2. We first divide by 5 and then take the square root. Remember that to find all solutions, when we take even roots, we must
consider both the positive and the negative roots. We have
5x2 = 32
x2 = 6.4
√
x = ± 6.4 = ±2.530.
3. This equation is linear in x. We have
4x + 1 = 8
4x = 7
x = 1.75.
4. We isolate the t5 , then raise both sides to the 1/5 power:
2t5 = 74
t5 = 37
t = 371/5 = 2.059.
248
Chapter Seven /SOLUTIONS
5. This equation is linear in p. We have
5p − 12 = 3p + 8
2p = 20
p = 10.
6. This equation is linear in t. We have
1.3t + 10.9 = 6.2
1.3t = −4.7
t = −3.615.
7. We isolate the square root, then square both sides. We have
√
25 t = 8
√
t = 0.32
t = 0.322 = 0.1024.
8. We isolate the w3 , then raise both sides to the 1/3 power. We have
8w3 + 5 = 30
8w3 = 25
w3 = 3.125
w = 3.1251/3 = 1.462.
9. We isolate the p4 , then raise both sides to the 1/4 power. Since we are taking an even root, we must include both the
positive and negative values. We have
1.2p4 = 60
p4 = 50
p = ±501/4 = ±2.659.
10. This is a linear equation. We have
5.2x − 17.1 = 3.9x + 15.8
1.3x = 32.9
x = 25.308.
11. We isolate the x2 , then take the square root of both sides. Since we are taking an even root, we must include both the
positive and negative values. We have
2.3x2 − 4.5 = 6.8
2.3x2 = 11.3
x2 = 4.913
√
x = ± 4.913 = ±2.217.
SOLUTIONS to Review Problems for Chapter Seven
12. We isolate the power t2.6 first. We then raise both sides to the 1/2.6 power. We have
25.6t2.6 = 83.1
t2.6 = 3.246
t = (3.246)1/2.6 = 1.573.
13. We collect the variables on the left and the constants on the right, then raise both sides to the 1/3 power:
6q 5 = 15q 2
q5
15
=
q2
6
q 3 = 2.5
q = (2.5)1/3 = 1.357.
14. We collect the variables on the left and the constants on the right, then raise both sides to the 1/1.3 power:
25.4x3.7 = 4.6x2.4
x3.7
4.6
=
x2.4
25.4
x1.3 = 0.1811
x = (0.1811)1/1.3 = 0.269.
15. We collect the variables on the left and the constants on the right, then raise both sides to the 1/2.2 power:
25t5.2 = 4.9t3
t5.2
4.9
=
t3
25
t2.2 = 0.196
t = (0.196)1/2.2 = 0.477.
16. We isolate the x3 , then raise both sides to the 1/3 power:
5(x3 − 31) = 27
5x3 − 155 = 27
5x3 = 182
x3 = 36.4
x = (36.4)1/3 = 3.314.
17. We isolate the t5 , then raise both sides to the 1/5 power:
3.1(t5 + 12.4) = 10
3.1t5 + 38.44 = 10
3.1t5 = −28.44
t5 = −9.174
t = (−9.174)1/5 = −1.558.
249
250
Chapter Seven /SOLUTIONS
18. We have
4(s + 3) − 7(s + 10) = 25
4s + 12 − 7s − 70 = 25
−3s − 58 = 25
−3s = 83
83
= −27.667.
s=−
3
19. We use the distributive law and then isolate the r 2 . When we take the square root, since we are taking an even root, we
must include both the positive and negative values. We have
10(r 2 − 4) = 18
10r 2 − 40 = 18
10r 2 = 58
r 2 = 5.8
√
r = ± 5.8 = ±2.408.
20. We have
5.2(t2 + 3.1) = 12
5.2t2 + 16.12 = 12
5.2t2 = −4.12
t2 = −0.792.
Since t2 is never negative, this equation has no solutions.
21. We isolate the x3 , then raise both sides to the 1/3 power:
ax3 = b
b
x3 =
a
x=
p
3
b/a.
22. Since we are solving for r, we put terms with r on one side of the equation, and terms without r on the other side. We
then raise both sides to the 1/3 power:
p2 r 3 − 5p3 = 100
p2 r 3 = 100 + 5p3
100 + 5p3
r3 =
p2
r=
100 + 5p3
p2
1/3
.
23. Since we are solving for p, we put terms with p on one side of the equation, and terms without p on the other side before
taking the square root of both sides:
5p2 + 6q + 17 = 8q
5p2 = 2q − 17
SOLUTIONS to Review Problems for Chapter Seven
251
2q − 17
= 0.4q − 3.4
p5
p = ± 0.4q − 3.4 if the solution exists.
p2 =
We must include both the positive and negative values because we are taking an even root.
24. This is a linear equation in q. We put terms with q on one side of the equation, and terms without q on the other side. We
have
5p2 + 6q + 17 = 8q
5p2 + 17 = 2q
5p2 + 17
= 2.5p2 + 8.5.
q=
2
25. This is a linear equation in y. We first use the distributive law, and then put terms with y on one side of the equation, and
terms without y on the other side. We have
5x(2x + 6y) = 2y(3x + 10)
10x2 + 30xy = 6xy + 20y
24xy − 20y = −10x2
(24x − 20)y = −10x2
y=
−10x2
.
24x − 20
26. We first use the distributive law, and then put terms with s on one side of the equation, and terms without s on the other
side. We then raise both sides to the 1/3 power:
6r 2 (s3 + 5rt + 2) = 8rt − 15
6r 2 s3 + 30r 3 t + 12r 2 = 8rt − 15
6r 2 s3 = 8rt − 30r 3 t − 12r 2 − 15
8rt − 30r 3 t − 12r 2 − 15
s3 =
6r 2
s=
8rt − 30r 3 t − 12r 2 − 15
6r 2
1/3
.
27. This is a linear equation in t. We first use the distributive law, and then put terms with t on one side of the equation, and
terms without t on the other side. We have
6r 2 (s3 + 5rt + 2) = 8rt − 15
2 3
6r s + 30r 3 t + 12r 2 = 8rt − 15
30r 3 t − 8rt = −6r 2 s3 − 12r 2 − 15
(30r 3 − 8r)t = −6r 2 s3 − 12r 2 − 15
t=
−6r 2 s3 − 12r 2 − 15
.
30r 3 − 8r
28. We first use the distributive law, and then put terms with x on one side of the equation, and terms without x on the other
side. We then square both sids of the equation:
√
6 xy − 32y 4 + 56y 2 = 3y(6y 3 + 5y)
√ √
6 x y − 32y 4 + 56y 2 = 18y 4 + 15y 2
252
Chapter Seven /SOLUTIONS
√ √
6 x y = 50y 4 − 41y 2
√
50y 4 − 41y 2
x=
√
6 y
x=
(50y 4 − 41y 2 )2
36y
with y > 0.
29. We isolate the cube root, cube both sides, and then solve for x. We have
√
3
A + Bx + C = D
√
3
Bx + C = D − A
Bx + C = (D − A)3
Bx = (D − A)3 − C
x=
(D − A)3 − C
.
B
30. We first use the distributive law. Then, since we are solving for S, we put terms with S on one side of the equation, and
terms without S on the other side and take the square root of both sides:
25V0 S 2 [T ] + 10(H 2 + V0 ) = A0 (3V + V0 )
25V0 S 2 [T ] + 10H 2 + 10V0 = 3A0 V + A0 V0
25V0 S 2 [T ] = 3A0 V + A0 V0 − 10H 2 − 10V0
S2 =
3A0 V + A0 V0 − 10H 2 − 10V0
25V0 [T ]
S=±
r
3A0 V + A0 V0 − 10H 2 − 10V0
if the solution exists.
25V0 [T ]
We must include both the positive and negative values because we are taking an even root.