556
Chapter 20
◆
Exponential and Logarithmic Functions
Find to four significant digits the number whose natural logarithm is the given value.
55.
58.
61.
64.
2.846
2.846
0.936
9.47
20–4
56.
59.
62.
65.
57.
60.
63.
66.
4.263
0.365
4.97
18.36
0.879
5.937
15.84
21.83
Properties of Logarithms
In this section we use the properties of logarithms along with our ability to convert back and
forth between exponential and logarithmic forms. These together will enable us to later solve
equations that we could not previously handle, such as exponential equations.
Products
We wish to write an expression for the log of the product of two positive numbers, say, M and N.
logb MN ?
Let us substitute: M bc and N bd. Then
MN bcbd bcd
Here, b 0 and b 1.
by the laws of exponents. Writing this expression in logarithmic form, we have
logb MN c d
But, since bc M, c logb M. Similarly, d logb N. Substituting, we have the following equation:
Log of a
Product
logb MN logb M logb N
187
The log of a product equals the sum of the logs of the factors.
◆◆◆
Example 30:
(a) log 7x log 7 log x
(b) log 3 log x log y log 3xy
◆◆◆
The log of a sum is not equal to the sum of the logs.
Common
Error
logb(M N) logb M logb N
The product of two logs is not equal to the sum of the logs.
(logb M)(logb N) logb M logb N
We don’t pretend that this
is a practical problem, now
that we have calculators. But
we recommend doing a few
logarithmic computations
just to get practice in using
the properties of logarithms,
which we’ll need later to solve
logarithmic and exponential
equations.
◆◆◆
Example 31: Multiply (2.947)(9.362) by logarithms.
Solution: We take the log of the product
log(2.947 9.362) log 2.947 log 9.362
0.469 38 0.971 37 1.440 75
This gives us the logarithm of the product. To get the product itself, we take the antilog.
(2.947)(9.362) 101.44075 27.59
◆◆◆
Section 20–4
◆
557
Properties of Logarithms
Quotients
Let us now consider the quotient M divided by N. Using the same substitution as above, we obtain
M bc
bcd
N
bd
Going to logarithmic form yields
M
logb c d
N
Finally, substituting back gives us the following equation:
Log of a
Quotient
M
logb logb M logb N
N
188
The log of a quotient equals the log of the dividend minus the log of the divisor.
◆◆◆
Example 32:
3
log log 3 log 4
4
◆◆◆
◆◆◆
Example 33: Express the following as the sum or difference of two or more logarithms:
ab
log xy
Solution: By Eq. 188,
ab
log log ab log xy
xy
and by Eq. 187,
log ab log xy log a log b (log x log y)
log a log b log x log y
◆◆◆
Example 34: The expression log 10x2 log 5x can be expressed as a single logarithm as
follows:
10x2
◆◆◆
log or
log 2x
5x
◆◆◆
Similar errors are made with quotients as with products.
Common
Error
◆◆◆
logb (M N ) logb M logb N
logb M
logb M logb N
logb N
Example 35: Divide using logarithms:
28.47 1.883
Solution: By Eq. 188,
28.47
log log 28.47 log 1.883
1.883
1.4544 0.2749 1.1795
558
Chapter 20
◆
Exponential and Logarithmic Functions
This gives us the logarithm of the quotient. To get the quotient itself, we take the antilog.
28.47 1.883 101.1795 15.12
◆◆◆
Powers
Consider now the quantity M raised to a power p. Substituting bc for M, as before, we have
Mp (bc)p bcp
In logarithmic form,
logb Mp cp
Substituting back gives us the following equation:
Log of a
Power
logb Mp p logb M
189
The log of a number raised to a power equals the power times the log of the number.
◆◆◆
Example 36:
log 3.851.4 1.4 log 3.85
◆◆◆
Example 37: The expression 7 log x can be expressed as a single logarithm with a coefficient of 1 as follows:
◆◆◆
log x7
◆◆◆
◆◆◆
Example 38: Express 2 log 5 3 log 2 4 log 1 as a single logarithm.
Solution:
2 log 5 3 log 2 4 log 1 log 52 log 23 log 14
25(8)
log log 200
1
◆◆◆
◆◆◆
Example 39: Express as a single logarithm with a coefficient of 1:
a
x
2
2 log 4 log 3 log y
3
b
Solution:
a
x
a4
x3
2
22
2 log 4 log 3 log log log log 2
4
y
3
b
3
b
y3
4 3
4a y
log 9b4x3
◆◆◆
◆◆◆
Example 40: Rewrite without logarithms the equation
log(a2 b2) 2 log a log(a b) 3 log b
Solution: We first try to combine all of the logarithms into a single logarithm. Rearranging gives
log(a2 b2) log(a b) 2 log a 3 log b 0
Section 20–4
◆
559
Properties of Logarithms
By the properties of logarithms,
a2 b2
a2
log log 0
ab
b3
or
a2(a b)(a b)
log 0
b3(a b)
Simplifying yields
a2 (a + b)
log 0
b3
We then rewrite without logarithms by going from logarithmic to exponential form.
a2 (a b)
100 1
b3
or
a2(a b) b3
◆◆◆
◆◆◆
Example 41: Find 35.821.4 by logarithms to four significant digits.
Solution: By Eq. 189,
log 35.821.4 1.4 log 35.82
1.4(1.5541) 2.1758
Taking the antilog yields
35.821.4 102.1758 149.9
◆◆◆
Roots
We can write a given radical expression in exponential form and then use the rule for powers.
Thus, by Eq. 36,
q
M M1/q
Taking the logarithm of both sides, we obtain
q
logb M logb M1/q
Then, by Eq. 189, the following equation applies:
Log of
a Root
q 1
logbM logb M
q
190
The log of the root of a number equals the log of the number divided by the index of the
root.
◆◆◆
Example 42:
(a) log 8 15 log 8 15 (0.9031) 0.1806
5
4 (b) log x log
z
2
4
log x
log z
◆◆◆
We may take the logarithm of
both sides of an equation, just
as we took the square root of
both sides of an equation or
took the sine of both sides of an
equation.
560
Chapter 20
◆◆◆
◆
Exponential and Logarithmic Functions
5
Example 43: Find 3824 by logarithms.
Solution: By Eq. 190,
3.5825
1
5 log 3824 log 3824 5
5
0.7165
Taking the antilog gives us
5
3824 100.7165 5.206
◆◆◆
Log of 1
Let us take the log to the base b of 1 and call it x. Thus
x logb 1
Switching to exponential form, we have
bx 1
This equation is satisfied, for any positive b, by x 0. Therefore:
logb 1 0
Log of 1
191
The log of 1 is zero.
Log of the Base
We now take the logb of its own base b. Let us call this quantity x.
logb b x
Going to exponential form gives us
bx b
So x must equal 1.
logb b 1
Log of the Base
192
The log (to the base b) of b is equal to 1.
◆◆◆
Example 44:
(a) log 1 0
(c) log5 5 1
(e) ln e 1
(b) ln 1 0
(d) log 10 1
◆◆◆
Log of the Base Raised to a Power
Consider the expression logb bn. We set this expression equal to x and, as before, go to exponential form.
logb bn x
bx bn
xn
Therefore:
Log of the Base
Raised to a Power
logb bn n
The log (to the base b) of b raised to a power is equal to the power.
193
Section 20–4
◆
561
Properties of Logarithms
◆◆◆
Example 45:
(a) log2 24.83 4.83
(b) loge e2x 2x
(c) log10 0.0001 log10 104 4
◆◆◆
Base Raised to a Logarithm of the Same Base
We want to evaluate an expression of the form
blogbx
Setting this expression equal to y and taking the logarithms of both sides, gives us
y blogbx
logb y logb blogb x logb x logb b
Since logb b 1, we have logb y logb x. Taking the antilog of both sides, we have
x y blogb x
Thus:
Base Raised to a
Logarithm of the
Same Base
blogbx x
194
◆◆◆
Example 46:
(a) 10logw w
(b) eloge3x 3x
◆◆◆
Change of Base
We can convert between natural logarithms and common logarithms (or between logarithms
to any base) by the following procedure. Suppose that log N is the common logarithm of some
number N. We set it equal to x.
log N x
Going to exponential form, we obtain
10x N
Now we take the natural log of both sides.
ln 10x ln N
By Eq. 189,
x ln 10 ln N
ln N
x ln 10
But x log N, so we have the following equation:
Change
of Base
ln N
log N ln 10
195
where ln 10 2.3026
The common logarithm of a number is equal to the natural log of that number divided by
the natural log of 10.
To answer Example 45(a) on
the Ti-89, press CATALOG,
select “log” from the menu and
press ENTER. Then type in the
numbers. The display will look
like this: log(2^.83, 2).
562
Chapter 20
◆◆◆
◆
Exponential and Logarithmic Functions
Example 47: Find log N if ln N 5.824.
Solution: By Eq. 195,
5.824
log N 2.529
2.3026
◆◆◆
◆◆◆
Example 48: What is ln N if log N 3.825?
Solution: By Eq. 195,
ln N ln 10(log N)
2.3026(3.825) 8.807
Exercise 4
◆
◆◆◆
Properties of Logarithms
Write as the sum or difference of two or more logarithms.
2
2. log 4x
1. log 3
x
3. log ab
4. log 2
5. log xyz
6. log 2ax
3x
7. log 4
1
9. log 2x
abc
11. log d
5
8. log xy
2x
10. log 3y
x
12. log 2ab
Express as a single logarithm with a coefficient of 1. Assume that the logarithms in each problem have the same base.
13. log 3 log 4
15. log 2 log 3 log 4
17. 4 log 2 3 log 3 2 log 4
14. log 7 log 5
16. 3 log 2
18. log x log y log z
log x log y
20. 3
2
19. 3 log a 2 log b 4 log c
y
21. log ax 2 log b 3 log cz
22. 2 log x log(a b) 12 log(a bx)
1
23. log(x 2) log(x 2)
2
Rewrite each equation so that it contains no logarithms.
24.
25.
26.
27.
28.
29.
30.
log x 3 log y 0
log2 x 2 log2 y x
2 log(x 1) 5 log(y 2)
log2(a2 b2) 1 2 log2 a
log(x y) log 2 log x log(x2 y2)
log(p2 q2) log(p q) 2
log2 2
31. loge e
2
33. log3 3
36. eloge
x
x
32. log10 10
34. loge e
35. log10 10x
37. 2log23y
38. 10logx
2
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
◆◆◆