Download 563 20–5 Exponential Equations

Section 20–5
◆
563
Exponential Equations
Logarithmic Computation
This may not seem like hot
stuff in the age of calculators
and computers, but the use
of logarithms revolutionized
computation in its time. The
French mathematician Laplace
wrote: “The method of logarithms,
by reducing to a few days
the labour of many months,
doubles, as it were, the life
of the astronomer, besides
freeing him from the errors and
disgust inseparable from long
calculation.”
Evaluate using logarithms.
39. 5.937 92.47
41. 3.97 8.25 9.82
9 43. 8563
45. 83.620.5720
7 47. 8364
40. 6923 0.003 846
42. 88.25 42.94
44. 4.8363.970
3 46. 587
4 48. 62.4
Change of Base
Find the common logarithm of the number whose natural logarithm is the given value.
49. 8.36
52. 15.36
50. 3.846
53. 5.26
51. 3.775
54. 0.638
Find the natural logarithm of the number whose common logarithm is the given value.
55. 84.9
58. 73.9
57. 3.82
60. 2.63
56. 2.476
59. 2.37
20–5 Exponential Equations
Solving Exponential Equations
We return now to the problem we started in Sec. 20–3 but could not finish, that of finding the
exponent when the other quantities in an exponential equation were known. We tried to solve
the equation
24.0 2.48x
The key to solving exponential equations is to take the logarithm of both sides. This enables us to
use Eq. 189 to extract the unknown from the exponent. Taking the logarithm of both sides gives
log 24.0 log 2.48x
By Eq. 189,
log 24.0 x log 2.48
log 24.0
1.380
x 3.50
log 2.48 0.3945
In this example we took common logarithms, but natural logarithms would have worked just
as well.
◆◆◆
Example 49: Solve for x to three significant digits:
3.25x2 1.443x1
Solution: We take the log of both sides. This time choosing to use natural logs, we get
(x 2) ln 3.25 (3x 1) ln 1.44
ln 3.25
(x 2) 3x 1
ln 1.44
By calculator,
3.232(x 2) 3x 1
Removing parentheses and collecting terms yields
0.232x 7.464
x 32.2
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564
Chapter 20
◆
Exponential and Logarithmic Functions
Solving Exponential Equations with Base e
If an exponential equation contains the base e, taking natural logs rather than common logarithms will simplify the work.
◆◆◆
Example 50: Solve for x:
157 112e3x2
Solution: Dividing by 112, we have
1.402 e3x2
Taking natural logs gives us
ln 1.402 ln e3x2
By Eq. 189,
ln 1.402 (3x 2) ln e
But ln e 1, so
ln 1.402 3x 2
ln 1.4022
x 0.554
3
◆◆◆
◆◆◆
Example 51: Solve for x to three significant digits:
3ex 2ex 4(ex ex)
Solution: Removing parentheses gives
3ex 2ex 4ex 4ex
Combining like terms yields
ex 6ex 0
Factoring gives
ex(e2x 6) 0
This equation will be satisfied if either of the two factors is equal to zero. So we set each factor
equal to zero (just as when solving a quadratic by factoring) and get
ex 0
and
e2x 6
There is no value of x that will make ex equal to zero, so we get no root from that equation.
We solve the other equation by taking the natural log of both sides.
2x ln e ln 6 1.792
x 0.896
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Our exponential equation might be of quadratic type, in which case it can be solved by
substitution as we did in Chapter 14.
◆◆◆
Example 52: Solve for x to three significant digits:
3e2x 2ex 5 0
Solution: If we substitute
w ex
our equation becomes the quadratic
3w2 2w 5 0
Section 20–5
◆
565
Exponential Equations
By the quadratic formula,
2 4 4(3)(5)
w 1
6
and
5
3
Substituting back, we have
ex 1
and
5
ex 3
So x ln(1), which we discard, and
5
x ln 0.511
3
◆◆◆
Solving Exponential Equations with Base 10
If an equation contains 10 as a base, the work will be simplified by taking common logarithms.
◆◆◆
Example 53: Solve 105x 2(102x) to three significant digits.
Solution: Taking common logs of both sides gives us
log 105x log[2(102x)]
By Eqs. 187 and 189,
5x log 10 log 2 2x log 10
By Eq. 192, log 10 1, so
5x log 2 2x
3x log 2
log 2
x 0.100
3
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Doubling Time
Being able to solve an exponential equation for the exponent allows us to return to the formulas
for exponential growth and decay (Sec. 20–2) and to derive two interesting quantities: doubling
time and half-life.
If a quantity grows exponentially according to the following function:
y aent
Exponential Growth
199
it will eventually double (y will be twice a). Setting y 2a, we get
2a aent
or 2 ent. Taking natural logs, we have
ln 2 ln ent nt ln e nt
also given in the following form:
Doubling Time
ln 2
t n
200
Since ln 2 0.693, if we let P
be the rate of growth expressed
as a percent (P 100n), not as
a decimal, we get a convenient
rule of thumb:
70
doubling time P
566
Chapter 20
◆
Exponential and Logarithmic Functions
◆◆◆ Example 54: In how many years will a quantity double if it grows at the rate of 5.0% per
year?
Solution:
ln 2
doubling time 0.050
0.693
0.050
14 years
◆◆◆
Half-Life
When a material decays exponentially according to the following function:
y aent
Exponential Decay
201
the time it takes for the material to be half gone is called the half-life. If we let y a/2,
1
2
1
e
ent nt
2 ent
Notice that the equation for halflife is the same as for doubling
time. The rule of thumb is, of
course, also the same. The
rule of thumb gives us another
valuable tool for estimation.
Taking natural logarithms gives us
ln 2 ln ent nt
ln 2
t n
Half-Life
200
◆◆◆
Example 55: Find the half-life of a radioactive material that decays exponentially at the
rate of 2.0% per year.
Solution:
ln 2
half-life 35 years
0.020
◆◆◆
Change of Base
If we have an exponential expression, such as 5x, we can convert it to another exponential
expression with base e or any other base.
◆◆◆
Example 56: Convert 5x into an exponential expression with base e.
Solution: We write 5 as e raised to some power, say, n.
5 en
Taking the natural logarithm, we obtain
ln 5 ln en n
So n ln 5 1.61, and 5 e1.61. Then
5x (e1.61)x e1.61x
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Section 20–5
◆◆◆
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567
Exponential Equations
Example 57: Convert 24x to an exponential expression with base e.
Solution: We first express 2 as e raised to some power.
2 en
ln 2 ln en n
n ln 2 0.693
Then
24x (e0.693)4x e2.773x
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Example 58: A bacteria sample contains 2850 bacteria and grows at a rate of 5.85% per
hour. Assuming that the bacteria grow exponentially:
(a) Write the equation for the number N of bacteria as a function of time.
(b) Find the time for the sample to grow to 10 000 bacteria.
Estimate: Note that the doubling time here is 70 5.85 or about 12 hours. To grow from 2850
to 10 000 will require nearly two doublings, or nearly 24 hours.
Solution:
(a) We substitute into the equation for exponential growth, y aent, with y N, a 2850,
and n 0.0585, getting
N 2850e0.0585t
(b) Setting N 10 000 and solving for t, we have
10 000 2850e0.0585t
e0.0585t 3.51
We take the natural log of both sides and set ln e 1.
0.0585t ln e ln 3.51
ln 3.51
t 21.5 h
0.0585
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which agrees with our estimate.
Exercise 5
◆
Exponential Equations
Solve for x to three significant digits.
1. 2x 7
2. (7.26)x 86.8
3. (1.15)x 2 12.5
4. (2.75)x (0.725)x
5. (15.4)
x
72.8
7. 5.62e3x 188
9. e2x 1 3ex 3
2
6. e5x 125
8. 1.05e4x 1 5.96
10. 14.8e3x 144
2
11. 52x 73x 2
12. 3x 175x 1
13. 103x 3(10x)
14. ex ex 2(ex ex)
15. 23x 1 32x 1
16. 52x 33x 1
17. 7e1.5x 2e2.4x
18. e4x 2e2x 3 0
19. ex ex 2
20. e6x e3x 2 0
2
Hint: Problems 18 through 20
are equations of quadratic type.
568
Chapter 20
◆
Exponential and Logarithmic Functions
Convert to an exponential expression with base e.
21. 3(43x)
22. 4(2.24x)
Applications
23. The current i in a certain circuit is given by
i 6.25e125t (amperes)
where t is the time in seconds. At what time will the current be 1.00 A?
24. The current through a charging capacitor is given by
E
i et/RC
R
(A83)
If E 325 V, R 1.35 , and C 3210 ␮F, find the time at which the current through
the capacitor is 0.0165 A.
25. The voltage across a charging capacitor is given by
v E(1 et/RC)
(A84)
If E 20.3 V and R 4510 , find the time when the voltage across a 545-␮F capacitor
is equal to 10.1 V.
–
26. The temperature above its surroundings of an iron casting initially at 1100 °C will be
T 11–00e0.0620t
after t minutes. Find the time for the casting to be at a temperature of 275 °C above its
surroundings.
27. A certain long pendulum, released from a height y0 above its rest position, will be at a
height
y y0e0.75t
at t seconds. If the pendulum is released at a height of 15 cm, at what time will the height
be 5.0 cm?
–
28. A population growing at a rate of 2.0% per year from an initial population of 9000 will
grow in t years to an amount
–
P 9000e0.02t
How many years will it take the population to triple?
29. The barometric pressure in kPa at a height of h metres above sea level is
p 99ekh
where k 1.25 104. At what height will the pressure be 33 kPa?
30. The approximate density of seawater at a depth of h km is
␪
d 1024e0.004 23h (kg/m3)
Find the depth at which the density will be 1032 kg/m3.
31. A rope passing over a rough cylindrical beam (Fig. 20–15) supports a weight W. The force
F needed to hold the weight is
W
FIGURE 20–15
F We␮␪
F
where ␮ is the coefficient of friction and ␪ is the angle of wrap in radians. If ␮ 0.150,
–
–
what angle of wrap is needed for a force of 100 lb. to hold a weight of 200 lb.?
Section 20–6
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569
Logarithmic Equations
32. Using the formula for compound interest, Eq. A10, calculate the number of years it will
take a sum of money to triple when invested at a rate of 12% per year.
33. Using the formula for present worth, given in problem 7 of Exercise 1, in how many years
will $50,000 accumulate to $70,000 at 15% interest?
34. Using the annuity formula, given in problem 10 of Exercise 1, find the number of years it
will take a worker to accumulate $100,000 with an annual payment of $3,000 if the interest
rate is 8.0%.
35. Using the capital recovery formula from problem 12 of Exercise 1, calculate the number of
years a person can withdraw $10,000/y from a retirement fund containing $60,000 if the
rate of interest is 6 34 %.
36. Find the half-life of a material that decays exponentially at the rate of 3.50% per year.
37. How long will it take the U.S. annual oil consumption to double if it is increasing exponentially at a rate of 7.0% per year?
38. How long will it take the world population to double at an exponential growth rate of
1.64% per year?
39. What is the maximum annual growth in energy consumption permissible if the consumption is not to double in the next 20 years?
20–6 Logarithmic Equations
Often a logarithmic expression can be evaluated, or an equation containing logarithms can be
solved, simply by rewriting it in exponential form.
◆◆◆
Example 59: Evaluate x log5 25.
Solution: Changing to exponential form, we have
5x 25 52
x2
Since 52 25, the answer checks.
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Example 60: Solve for x:
1
logx 4 2
Solution: Going to exponential form gives us
x1/2 4
Squaring both sides yields
x 16
Since 161/2 4, the answer checks.
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Example 61: Solve for x:
3
log25 x 2