Download CHAPTER SEVEN

7.1 SOLUTIONS
57
CHAPTER SEVEN
Solutions for Section 7.1
EXERCISES
1. Since A = 1 · x2 , the exponent is 2 and the constant of proportionality is 1.
5. The power of r is 2 and the constant of proportionality is 4π.
9. p > 1
13. p > 1
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
21. As x increases, x4 increases. So if a > b, then a4 > b4 .
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.
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.
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 .
58
Chapter Seven /SOLUTIONS
r (cm)
2
√
r = 0.620 3 V
33.5
V (cm3 )
Figure 7.1
Solutions for Section 7.2
EXERCISES
1. This is a power function.
5. This is a power function. We can rewrite it as y = 12 x3 .
4
4
= 1/4 = 4z −1/4 . The coefficient is 4 and the exponent is −1/4.
9. We have √
4
z
z
13. We have
r
3
8
81/3
2
=
= 2 = 2x−2 . The coefficient is 2 and the exponent is −2.
1/3
6
x6
x
(x )
17. We have
2
√ = (2/3)x−1/2 ,
3 x
so k = 2/3 and p = −1/2.
21. We have
(3x2 )3 = 33 (x2 )3 = 27x6 .
Thus k = 27 and p = 6.
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 .
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.
33.
r
so k =
√
√
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).
7.3 SOLUTIONS
59
PROBLEMS
37. This is a power function with exponent 1 so the graph is a line and graph (III) 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.
45. We see that W is a power function of L with exponent 3, so graph (I) 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.
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.
57. Increases, since (ax)2 + a2 = a2 (x2 + 1).
61. Decreases, since
x + a−1 = x +
1
.
a
65. Decreases, since
(ax)1/3
a1/3 x1/3
x1/3
√
=
= 1/6 .
1/2
a
a
a
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.
Solutions for Section 7.3
EXERCISES
1. We raise both sides to the 1/3 power to obtain:
x = 501/3 = 3.684.
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
60
Chapter Seven /SOLUTIONS
p
9.
y − 2 = 11
y − 2 = 121
y = 123.
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.
17.
1
=0
L2
1
16 = 2
L
16L2 = 1
1
L2 =
16
1
L=± .
4
16 −
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
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.
29. (iv), because positive numbers have two square roots.
33. (vi), because even powers of real numbers are not negative.
37. (i), because the square root of x is positive only if x is positive.
41. (ii), because only negatives have negative odd powers.
7.3 SOLUTIONS
61
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
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.
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,
|
{z
√
8−x
}
making the value of the left-hand side less than 5.
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
M =
= 83.986 kg.
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
1095
S
3/2
=
1 3/2
1095
· S 3/2 = 0.0000276S 3/2 .
M = 0.0000276S 3/2 .
62
Chapter Seven /SOLUTIONS
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%.
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.
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.
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.
73. (a) If x = 1 then A = −4.
(b) If x > 1 then A < −4.
(c) This equation has no solution for A > 0.
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 .
7.4 SOLUTIONS
63
√
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.
9. For some constant k, we have S = kh2 .
13. When x is doubled, we have
So y is multiplied by a factor of 8.
new value of y = k(2x)3 = k · 23 x3 = 8(kx3 ).
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.
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.
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.
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.
64
Chapter Seven /SOLUTIONS
(c) When the side is 12 x,
Volume =
1
x
2
3
=
3
1
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.
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
mammal lifespan
10
20
30
Figure 7.2
(c)
(i) From LM = 11.8W 0.20 we find W =
LB = 28.3
L
M
0.19/0.20
LM
11.8
1/0.20
which, when substituted in LB = 28.3W 0.19 , gives
. 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.
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 to Review Problems for Chapter Seven
Solutions for Chapter 7 Review
EXERCISES
1. We have y = 3x−2 ; k = 3, p = −2.
5. We have y = 25 x−1/2 ; k = 52 , p = −1/2.
9. We have y = 53 · x3 = 125x3 ; k = 125, p = 3.
−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
17. We know that (4x3 )(3x−2 ) = 12x. Therefore, the base is x, the exponent is 1, and the coefficient is 12.
21. We have 3(−4r)2 = 3(16r 2 ) = 48r 2 . The base is r, the exponent is 2, and the coefficient is 48.
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
50
x3 =
2.8
1/3
50
= 2.614.
x=
2.8
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.
33. Since we take an even root to solve the equation, the equation has two solutions.
37. Since we take an odd root to solve the equation, this equation has only one solution.
PROBLEMS
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
=
The quantity R is inversely proportional to the cube root of P .
k1/3
Constant
=
.
P 1/3
P 1/3
65
66
Chapter Seven /SOLUTIONS
45. (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.
53. (i), because for a negative power of x to be less than 1, x must be greater than 1.
57. Taking the square root of both sides of the equation, we get
√
2x = ± 16 = ±4
x = ±2.
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.
Solutions for Solving Drill
1. We raise both sides to the 1/3 power:
x3 = 10
x = 101/3 = 2.154.
5. This equation is linear in p. We have
5p − 12 = 3p + 8
2p = 20
p = 10.
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.
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.
SOLUTIONS to Review Problems for Chapter Seven
67
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.
21. We isolate the x3 , then raise both sides to the 1/3 power:
ax3 = b
b
x3 =
a
x=
p
3
b/a.
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
29. We isolate the cube root, cube both sides, and then solve for x. We have
√
A + 3 Bx + C = D
√
3
Bx + C = D − A
Bx + C = (D − A)3
Bx = (D − A)3 − C
x=
(D − A)3 − C
.
B