20
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:
logbx = n means bn = x.
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 |
. |
| Answer. log4 | 1 16 |
= −2. |
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.
logb1 = 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.
logbb = 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:
logbbx = x
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. |
| log3 | 1 9 |
= | log33−2 = −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.
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| 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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