19. (A) If F Is Linear, Then the Formula for F(X) Is of the Form

Section 9.6

15. The function y = e-x will approach zero faster. To see this, note that a doubling of x in the cubic function y = x-3 will cause the y-value to decrease by a factor of 2-3 = 1/8; while a doubling of x in the exponential function y = e-x will cause the y-value to be squared. For small values of y, squaring decreases faster than multiplying by 2-3. To see this

numerically, look in Table 9.9.

19. (a) If f is linear, then the formula for f(x) is of the form

f(x) = mx + b,

where

Thus, if f is a linear function,

(b) If f is exponential, then the formula for f(x) is of the form

f(x) = abx, b > 0, b 6 =/ 1.

Taking the ratio of f(2) to f(−1), we have

and

Thus

b3 = 64

and

b = 4

To solve for a, note that

f(2) = a(4)2 = 48,

which gives a = 3. Thus, an exponential model for f is f(x) = 3 · 4x

f(x) = kxp, k and p constants.

Taking the ratio of f(2) to f(−1), we have

(-2)p = 64

Thus, p = 6. To solve for k, note that

f(2) = k · 26 = 48

which gives

64k = 48

Thus, f(x) = (3/4)x6 is a power function which satisfies the given data.

21.

A is kx5/7, B is kx9/16, C is kx3/8, D is kx3/11.

22. If f(x) = mx1/3 goes through (1, 2), then m(1)1/3 = 2, so m = 2 and f(x) = 2x1/3. Using x = 8 in f(x) = 2x1/3 gives t = 4. If g(x) = kx4/3 goes through (8, 4), then k = ¼. Thus, m = 2, t = 4, and k = ¼.

25. Since the denominator has highest power of t6, which dominates t2, the ratio tends to 0. Thus, y → 0 as t → ∞

or t → −∞.

30. Since ex dominates x100, for large positive x, the value of y is very large. Thus, y → ∞as x → ∞.

For large negative x, the value of ex is very small, but x100 is very large. Thus, y → 0 as x → −∞.

31. Since e3t dominates e3t, the value of y is very small for large positive t. Thus, y → 0 as t → ∞.

For large negative t, the value of e2t → 0, so y → 0 as t → −∞.

33. Since et and t2 both dominate ln |t|, we have y → ∞as t → ∞. For large negative t, the value of et → 0, but t2 is large and dominates ln |t|. Thus, y → ∞as t → −∞.

38. (a)

(b) The balance is not growing at a linear rate because the change (increase) in balance is increasing each year. If the

growth were linear, the increase would be the same amount each year.

The balance is not growing exponentially, either, because the ratio of successive balance amounts is not constant.

For example,

and

Thus, neither a linear nor an exponential function represents the growth of the balance in this situation.

Section 9.7

3. Regression on a calculator returns the exponential function h(x) = 2.35(1.44)x

5. (a) Regression on a calculator returns the power function f(x) = 201.353x2.111, where f(x) represents the total dry

weight (in grams) of a tree having an x cm diameter at breast height.

(b) Using our regression function, we obtain f(20) = 201.535(20)2.111 = 112,313.62 gm.

13. (a) The FM band appears linear, because the FM frequency always increases by 4 as the distance increases by 10. See Figure 9.58.

(b) The AM band is increasing at an increasing rate. These data could therefore represent an exponential relation.

frequency, 4, compared to the change in length, 10, then m = 4/10 = 0.4 So

y = b + 0.4x.

But the table tells us that f(5) = b + 0.4(5) = b + 2 = 88. Therefore, b = 86, and

y = 86 + 0.4x.

We could have also used a calculator or computer to determine the coefficients for the linear regression.

(d) Since the data for the AMband appear exponential, we wish to plot the natural log of the frequency against the length. Table 9.11 gives the values of the AM station numbers, y, and the natural log, ln y, of those station numbers as a function of their location on the dial, x.

The (x, ln y) data are very close to linear. Regression gives the coefficients for the linear equation ln y = b+mx,

yielding

ln y = 3.839 + 0.023x.

Solving for y gives:

elny = e3.839 + 0.023x

y = e3.839 + 0.023x (since eln y = y).

y = e3.839e 0.023x (since ax+y = axay)

Since e3.839 = 46.5

y = 46.5e0.023x.

14. (a) The function y = −83.039 + 61.514x gives a superb fit, with correlation coefficient r = 0.99997.

(b) When the power function is plotted for 2 ≤ x ≤ 2.05, it resembles a line. This is true for most of the functions we

have studied. If you zoom in close enough on any given point, the function begins to resemble a line. However, for

other values of x (say, x = 3, 4, 5 . . .), the fit no longer holds.

19. (a) See Figure 9.62.

(b) Using a calculator or computer, we get P(t) = 56.108(1.031)t . Answers may vary.

50,400. The growth factor of 1.031 means the rate of growth is approximately 3.1% per year.

(d) We find P(100) = 1194.308, which is slightly higher than the given data value of 1,170.8.

(e) The estimated population, P(150) = 5510.118, is higher than the given census population.

20. (a) P(t) = 3.956(1.030)t . Answers may vary.

(b) The value, P(10) = 3.956 · 1.03010 = 5.314 for 1800, is an interpolation, while in the other problem, 1800 was

outside the data set of 1650-1790 and was a less accurate extrapolation.

(c) P(210) = 3.956·1.030210 = 1963.692. This is almost 2 billion, and far from being correct, since the US population was under 300 million in 2000.

27. (a) Quadratic is the only choice that increases and then decreases to match the data.

(b) Using ages of x = 20, 30, . . . , 80, a quadratic function is y = −31.798x2 + 3218.719x − 37,169.5. Answers may vary.

(d) The value of the function for age 10 is y = −31.798 · 102 + 3218.719 · 10 − 37,169.5 = −$8162. Answers may

vary. Not reasonable, as income is positive. In addition, 10-year olds do not usually work.

Section 4.1

1. The statement is equivalent to 19 = 10 1.279.

3. The statement is equivalent to 26 = e3.258

9. The statement is equivalent to v = log α.

11. We are solving for an exponent, so we use logarithms. We can use either the common logarithm or the natural logarithm. Since 23 = 8 and 24 = 16, we know that x must be between 3 and 4. Using the log rules, we have

2x = 11

log(2x) = log(11)

x log(2) = log(11)

If we had used the natural logarithm, we would have

15. We begin by dividing both sides by 17 to isolate the exponent:

We then take the log of both sides and use the rules of logs to solve for w:

18. We begin by dividing both sides by 25 to isolate the exponent:

20 = (1.1)3x.

We then take the log of both sides and use the rules of logs to solve for x:

log 20 = log(1.1)3x

log 20 = 3x log(1.1)

19. To do these problems, keep in mind that we are looking for a power of 10. For example, log 10,000 is asking for the power of 10 which will give 10,000. Since 104 = 10,000, we know that log 10,000 = 4.

(a) Since 1 = 100, log 1 = 0.

(b) Since 0.1 = 10-1, we know that log 0.1 = log 10-1 = -1.

similar to what we saw in (a), that log 1 = 0.

(d) To find the log√10 we need to recall that √10 = 101/2. Now we can use our identity and say log√10 =

log 101/2 = ½.

(e) Using the identity, we get log 105 = 5.

(f) Using the identity, we get log 102 = 2

For the last three problems, we’ll use the identity 10log N = N.

(h) 10log100 = 100

(i) 10log1 = 1

(j) 10log0.01 = 0.01

25. (a) Using the identity ln eN = N , we get ln e2x = 2x.

(b) Using the identity eln N = N, we get eln(3x+2) = 3x + 2.

29. (a) Since p = logm, we have m = 10P.

(b) Since q = log n, we have n = 10q, and so

n3 = (10q)3 = 103q.

log(mn3) = log(10p · 103q)

= log(10p+3q)

Using the identity log 10N = N, we have

log(mn3) = p+3q

(d) Since √m = m1/2,

log√m = logm1/2

Using the identity log ab = b · log a we have

log m1/2 = ½log m

Since p = log m

33. To find a formula for S, we find the points labeled (x0, y1) and (x1, y0) in Figure 4.1. We see that x0 = 4 and that y1 = 27. From the graph of R, we see that

y0 = R(4) = 5.1403(1.1169)4 = 8.

To find x1 we use the fact that R(x1) = 27:

We have S(4) = 27 and S(15) = 8. Using the ratio method, we have

Now we can solve for a:

so S(x) = 42.0207(0.8953)x

35. Using the log rules, we have

Checking the answer with a calculator, we get

and we see that

37. First, we isolate the power on one side of the equation:

Taking the log of both sides of the equation gives

47.

44e0.15t = 50(1.2)t

ln(44e0.15t) = ln(50(1.2)t)

ln 44 + ln e0.15t = ln 50 + ln 1.2t

ln 44 + 0.15t = ln 50 + t ln 1.2

0.15t − (ln 1.2)t = ln 50 − ln 44

(0.15 − ln 1.2)t = ln 50 − ln 44

48. Using log a − log b = log (a/b) we can rewrite the left side of the equation to read

This logarithmic equation can be rewritten as

since if log a = b then 10b = a. Multiplying both sides of the equation by (1 + x) yields

102 (1 + x) = 1 − x

100 + 100x = 1 − x

101x = −99

Check your answer:

49. We have log(2x + 5) · log(9x2) = 0

In order for this product to equal zero, we know that one or both terms must be equal to zero. Thus, we will set each of

the factors equal to zero to determine the values of x for which the factors will equal zero. We have

Checking and substituting back into the original equation, we see that the three solutions work. Thus our solutions are

x = −2, 1/3, or -1/3.

Section 4.2

5. Writing this as Q = 12.1(10-0.11)t, we have a = 12.1, b = 10-0.11 = 0.7762, r = b − 1 = −22.38%, and k = ln b = −25.32%.

8. We can use exponent rules to place this in the form abt:

2-(t-5)/3 = 2-(1/3)(t-5)

= 25/3-(1/3)t

= 25/3 · 2-(1/3)t

= 25/3 · (2-1/3)t,

so a = 25/3 = 3.1748 and b = 2(-1/3) = 0.7937. To find k and r, we use the fact that:

r = b − 1 = −0.2063 = −20.63%

k = ln b = −0.2310 = −23.10%.

11. To convert to the form Q = aekt, we first say that the right sides of the two equations equal each other (since each equals Q), and then we solve for a and k. Thus, we have aekt = 4 · 81.3t. At t = 0, we can solve for a:

aek0 = 4 · 80

a · 1 = 4 · 1

a = 4.

Thus, we have 4ekt = 4 · 81.3t, and we solve for k:

Therefore, the equation is Q = 4e2.703t.

21. Let t be the doubling time, then the population is 2P0 at time t, so

2P0 = P0e0.2t

2 = e0.2t

0.2t = ln 2

24.

29.

31.