20
LOGARITHMS
Definition
Common logarithms
Natural logarithms
The three laws of logarithms
Proof of the laws of logarithms
A LOGARITHM is an exponent.
Since
23 = 8,
then 3 is called the logarithm of 8 with base 2. We write
3 = log28.
3 is the exponent to which 2 must be raised to produce 8.
We write the base 2 as a subscript.
Thus a logarithm is the exponent to which the base must be raised to produce a given number.
Since
104 = 10,000
then
log1010,000 = 4
"The logarithm of 10,000 with base 10 is 4."
4 is the exponent to which 10 must be raised to produce 10,000.
"104 = 10,000" is called the exponential form.
"log1010,000 = 4" is called the logarithmic form.
Here is the definition:
That base with that exponent produces x.
Example 1. Write in exponential form: log232 = 5
Answer. 25 = 32
Example 2. Write in logarithmic form: 4−2 = |
1 16 |
. |
Example 3. Evaluate log81.
Answer. 8 to what exponent produces 1? 80 = 1.
log81 = 0.
We can observe that in any base, the logarithm of 1 is 0.
Example 4. Evaluate log55.
Answer. 5 with what exponent will produce 5? 51 = 5.
log55 = 1.
In any base, the logarithm of the base itself is 1.
Example 5. log22m = ?
Answer. 2 raised to what exponent will produce 2m ? m, obviously.
log22m = m.
The following is an important formal rule, valid for any base b:
This rule embodies the very meaning of a logarithm. x -- on the right -- is the exponent to which the base b must be raised to produce bx.
The rule also shows that the inverse of the function logbx is the exponential function bx. We will see this in the following Topic.
Example 6 . Evaluate log3 |
1 9 |
. |
Answer. |
1 9 |
is equal to 3 with what exponent? |
1 9 |
= 3−2. |
Compare the previous rule.
Which numbers, then, will have negative logarithms?
Proper fractions
Example 7. log2 .25 = ?
Answer. .25 = ¼ = 2−2. Therefore,
log2 .25 = log22−2 = −2.
Example 8. log3 = ?
Answer. = 31/5. (Definition of a rational exponent.) Therefore,
log3 = log331/5 = 1/5.
Problem 1. Write each of the following in logarithmic form.
To see the answer, pass your mouse over the colored area. To cover the answer again, click "Refresh" ("Reload").
a) bn = x |
logbx = n |
|
b) 23 = 8 |
log28 = 3 |
|
c) 102 = 100 |
log10100 = 2 |
|
d) 5−2 = 1/25. |
log51/25 = −2. |
Problem 2. Write each of the following in exponential form.
a) logbx = n |
bn = x |
|
b) log232 = 5 |
25 = 32 |
|
c) 2 = log864 |
82 = 64 |
|
d) log61/36 = −2 |
6−2 = 1/36 |
Problem 3. Evaluate the following.
a) log216 |
= 4 |
|
b) log416 |
= 2 |
|
c) log5125 |
= 3 |
|
d) log81 |
= 0 |
|
e) log88 |
= 1 |
|
f) log101 |
= 0 |
Problem 4. What number is n?
a) log10n = 3 |
1000 |
|
b) 5 = log2n |
32 |
|
c) log2n = 0 |
1 |
|
d) 1 = log10n |
10 |
e) logn |
1 16 |
= −2 |
4 |
|
f) logn |
1 5 |
= −1 |
5 |
|
g) log2 |
1 32 |
= n |
−5 |
|
h) log2 |
1 2 |
= n |
−1 |
Problem 5. logbbx =
x
Problem 6. Evaluate the following.
a) log9 |
1 9 |
|
= log99−1 = −1 |
b) log9 |
1 81 |
|
= −2 |
|
c) log2 |
1 4 |
|
= −2 |
|
d) log2 |
1 8 |
|
= −3 |
|
e) log2 |
1 16 |
|
= −4 |
f) log10.01 |
−2 |
|
g) log10.001 |
−3 |
|
h) log6 |
= 1/3 |
|
i) logb |
= 3/4 |
Common logarithms
The system of common logarithms has 10 as its base. When the base is not indicated,
log 100 = 2
then the system of common logarithms -- base 10 -- is implied.
Here are the powers of 10 and their logarithms:
Powers of 10: |
|
1 1000 |
|
1 100 |
|
1 10 |
|
1 |
|
10 |
|
100 |
|
1000 |
|
10,000 |
|
Logarithms: |
|
−3 |
|
−2 |
|
−1 |
|
0 |
|
1 |
|
2 |
|
3 |
|
4 |
Logarithms replace a geometric series with an arithmetic series.
Problem 7. log 10n = ?
n. The base is 10.
Problem 8. log 58 = 1.7634. Therefore, 101.7634 = ?
58. 1.7634 is the common logarithm of 58. When 10 is raised to that exponent, 58 is produced.
Problem 9. log (log x) = 1. What number is x?
log a = 1, implies a = 10. (See above.) Therefore, log (log x) = 1 implies log x = 10. Since 10 is the base,
x = 1010 = 10,000,000,000
Natural logarithms
The system of natural logarithms has the number called "e" as its base. (e is named after the 18th century Swiss mathematician, Leonhard Euler.) e is the base used in calculus. It is called the "natural" base because of certain technical considerations.
ex has the simplest derivative. See Lesson 14 of An Approach to Calculus.)
e can be calculated from the following series involving factorials:
e |
= 1 + |
1 1! |
+ |
1 2! |
+ |
1 3! |
+ |
1 4! |
+ . . . |
e is an irrational number, whose decimal value is approximately
2.71828182845904.
To indicate the natural logarithm of a number, we use the notation "ln."
ln x means logex.
Problem 10. What number is ln e ?
ln e = 1. The logarithm of the base itself is always 1. e is the base.
Problem 11. Write in exponential form (Example 1): y = ln x.
ey = x.
e is the base.
The three laws of logarithms
1. logbxy = logbx + logby
"The logarithm of a product is equal to the sum of the logarithms of each factor."
2. logb |
x y |
= logbx − logby |
"The logarithm of a quotient is equal to the logarithm of the numerator minus the logarithm of the denominator."
3. logb x n = n logbx
"The logarithm of a power of x is equal to the exponent of that power times the logarithm of x."
We will prove these laws below.
Example 1. Use the laws of logarithms to rewrite log |
z5 |
. |
Answer. According to the first two laws,
log |
z5 |
= log x + log − log z5 |
Now, = y½. Therefore, according to the third law,
log |
z5 |
= log x + ½ log y − 5 log z |
Example 2. Use the laws of logarithms to rewrite ln .
Solution.
ln |
= |
ln (sin x ln x)½ |
|
|
= |
½ ln (sin x ln x), 3rd Law |
|
|
= |
½ (ln sin x + ln ln x), 1st Law |
Note that the factors sin x ln x are the arguments of the logarithm function.
Example 3. Solve this equation for x:
log 32x + 5 |
= |
1 |
|
Solution. According to the 3rd Law, we may write |
|
(2x + 5)log 3 |
= |
1 |
|
Now, log 3 is simply a number. Therefore, on multiplying out by log 3, |
|
2x· log 3 + 5 log 3 |
= |
1 |
|
2x· log 3 |
= |
1 − 5 log 3 |
|
x |
= |
1 − 5 log 3 2 log 3 |
By this technique, we can solve equations in which the unknown appears in the exponent.
Problem 12. Use the laws of logarithms to rewrite the following.
a) log |
ab c |
|
= log a + log b − log c |
|
b) log |
ab² c4 |
|
= log a + 2 log b − 4 log c |
|
c) log |
z |
|
= 1/3 log x + 1/2 log y − log z |
d) ln (sin²x ln x)
= ln sin²x + ln ln x = 2 ln sin x + ln ln x
e) ln |
= ½ ln (cos x· x1/3 ln x) |
|
|
= ½ (ln cos x + 1/3 ln x + ln ln x) |
f) ln (a2x − 1 b5x + 1 ) |
= ln a2x − 1 + ln b5x + 1 |
|
|
= (2x − 1) ln a + (5x + 1) ln b |
Problem 13. Solve for x.
ln 23x + 1 |
= |
5 |
|
(3x + 1) ln 2 |
= |
5 |
|
3x ln 2 + ln 2 |
= |
5 |
|
3x ln 2 |
= |
5 − ln 2 |
|
x |
= |
5 − ln 2 3 ln 2 |
Problem 14. Prove: −ln x |
= ln |
1 x |
. |
−ln x = (−1)ln x = ln
x−1 = ln |
1 x |
Proof of the laws of logarithms
The laws of logarithms will be valid for any base. We will prove them for base e, that is, for y = ln x.
1. ln ab = ln a + ln b.
The function y = ln x is defined for all positive real numbers x. Therefore there are real numbers p and q such that
p = ln a and q = ln b.
This implies
a = e p and b = e q.
Therefore, according to the rules of exponents,
ab = e p· e q = ep + q.
And therefore
ln ab = ln ep + q = p + q = ln a + ln b.
That is what we wanted to prove.
In a similar manner we can prove the 2nd law. Here is the 3rd:
3. ln an = n ln a.
There is a real number p such that
p = ln a;
that is,
a = e p.
And the rules of exponents are valid for all rational numbers n. (Lesson 29 of Algebra; an irrational number is the limit of a sequence of rational numbers.) Therefore,
an = e pn.
This implies
ln an = ln e pn = pn = np = n ln a.
That is what we wanted to prove.
Next Topic: Logarithmic and exponential functions
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