1. Introduction
Let a and N be positive real numbers and let N = an. Then n is called the logarithm of N to the base a. We write this as:
| n = logaN |
Example 1:
Exercise 1: Use the definition of logarithms given above to determine the value of x in each of the following:
Solution:
By the definition of a logarithm, we have:
But 27 = 33, so we have:
So:
Solution:
By the definition of a logarithm, we have:
Now:
So:
From this we see that x = ½
Solution: By the definition of a logarithm, we have:
Thus x = −2
Solution: By the definition of a logarithm, we have:
Thus x = 4
Solution: By the definition of a logarithm, we have:
Thus x = 8
Click on questions to reveal solutions
2. Rules of Logarithms
Let a, M, N be positive real numbers and k be any number. Then the following important rules apply to logarithms:
| Rule 1 | loga(MN) | = | logaM + logaN |
|---|---|---|---|
| Rule 2 | loga(MN) | = | logaM − logaN |
| Rule 3 | loga(mk) | = | k logaM |
| Rule 4 | logaa | = | 1 |
| Rule 5 | loga1 | = | 0 |
3. Logarithm of a Product
Proof that loga MN = loga M + loga N:
Let m = loga M and n = loga N, so, by definition, M = am and N = an. Then:
- where we have used the appropriate rule for exponents. From this, using the definition of a logarithm, we have:
But m + n = loga M + loga N, and the above equation may be written:
- which is what we wanted to prove.
Example 2:
If x = log6 36, then 6x = 36 = 62. Thus, log6 4 + log6 9 = 2.
Now 20 × 1 ⁄ 4 = 5, so log5 20 + log4 (1 ⁄ 4) = log5 5 = 1
Quiz 1: To which of the following numbers does the expression log315 + log30.6 simplify?
Solution:
Using Rule 1 we have: log3 15 + log3 0.6 = log3 (15 × 0.6) = log3 9
But, 9 = 32, so log3 15 + log30.6 = log332 = 2.
4. Logarithm of a Quotient
Proof that logaMN = loga M − loga N:
As before, let m = loga M and n = loga N. Then, M = am and N = an. Now we have:
- where we have used the appropriate rule for indices. By the definition of a logarithm, we have:
From this we are able to deduce that:
Example 3:
If x = log28, then 2x = 8 = 23 , so x = 3.
Since 3 ⁄ 5 = 0.6, then log3 0.6 = log3 (35) = log3 3 − log3 5.
Now log3 3 = 1, so that log3 0.6 = 1 − 1.465 = −0.465
Quiz 2: To which of the following numbers does the expression log212 − log2(3 ⁄ 4) simplify?
Solution:
Using Rule 2 we have: log212 − log2 (3 ⁄ 4) = log2 (12 ÷ 3)
Now we have 12 ÷ (3 ⁄ 4) = 12 × (4 ⁄ 3) = 12 × 43 = 16.
Thus log2 12 − log2 (3 ⁄ 4) = log2 16 = log224.
If x = log224, then 2x = 24 , so x = 4.
5. Logarithm of a Power
Proof that loga (Mk) = k loga M:
Let m = loga M, so M = am. Then:
- where we have used the appropriate rule for indices. From this we have, by the definition of a logarithm:
But m = loga M, so the last equation may be written:
- which is the result we wanted.
Example 4:
We have 10000 = 104, so 1 ⁄ 10000 = 1 ⁄ 104 = 10−4.
Thus, log10 (1 ⁄ 10000) = log10 (10−4) = −4 log10 10 = −4, where we have used Rule 4 to write log10 10 = 1.
We have 6 = √36 = 361 ⁄ 2 .
Thus, log36 6 = log36 (361 ⁄ 2) = 12 log36 36 = 12.
Quiz 3: If log3 5 = 1.465, which of the following numbers is log3 0.04?
Solution:
Note that: 0.04 = 4 ⁄ 100 = 1 ⁄ 25 = 1 ⁄ 52 = 5−2.
Thus, log3 0.04 = log3 (5−2) = −2 log3 5.
Since log3 5 = 1.465, we have log3 0.05 = −2 × 1.465 = −2.930
6. Use of the Rules of Logarithms
In this section we look at some applications of the rules of logarithms.
Example 5:
= log10 (103) = 3 log10 10 = 3.
= log10 100 = log10 (102) = 2 log10 10 = 2.
= loga (43 × 14) − loga (24) = loga (42) − loga (24)
= loga 16 − loga 16 = 0.
Exercise 2: Use the rules of logarithms to simplify each of the following:
Solution:
First of all, by Rule 3, we have 3 log3 2 = log3 (23) = log3 8. Thus, the expression becomes:
Using Rule 1, the term in the square brackets becomes:
The expression then simplifies to:
Solution:
First we use Rule 3 to rewrite the first two terms:
and:
Thus:
- where we have used Rule 1 to obtain the right hand side. Thus:
and, using Rule 2, this simplifies to:
Solution:
Dealing first with the expression in brackets, we have:
- where we have used, in succession, Rules 3 and 2. Now:
so that finally, we have:
| 2 loga 6 − (loga 4 + 2 loga 3) | = | loga (62) − loga (4 × 32) |
| = | loga (624 × 32) | |
| = | loga 1 | |
| = | 0 |
Solution:
Dealing first with the expression in brackets, we have:
- where we have used Rule 3 first, and then Rule 1. Now, using Rule 3 on the first term, followed by Rule 2, we obtain:
| 5 log3 6 − (2 log3 4 + log3 18) | = | log3 (65) − log3 (42 × 18) |
| = | log3 (6542 × 18) | |
| = | log3 (25 × 3542 × 2 × 9) | |
| = | log3(33) | |
| = | 3 log3 3 = 3 |
- since log3 3 = 1
Solution:
The first thing we note is that √3 can be written as 3½. We first simplify some of the terms. They are:
and
Putting all of this together:
| 3 log4 ( √3) − 12 log4 2 + 3 log4 2 − log4 6 | = | 32 log4 3 − 12 log43 + 3 log4 2 − (log4 2 + log4 3) |
| = | (32 − 12 − 1) log4 3 + (3 − 1) log4 2 | |
| = | 2 log3 2 = log4 (22) = log4 4 = 1 |
Click on questions to reveal solutions
7. Logarithms Quiz
In each of the following, find x
8. Change of Bases
There is one other rule for logarithms which is extremely useful in practice. This relates logarithms in one base to logarithms in a different base. Most calculators will have, as standard, a facility for finding logarithms to the base 10 and also for logarithms to base e (natural logarithms). What happens if a logarithm to a different base, for example 2, is required? The following is the rule that is needed:
| logac = logab × logbc |
Proof of the above rule:
Let x = loga b and y = logb c. Then, by the definition of logarithms:
This means that:
- with the last equality following from the laws of indices. Since c = axy , by the definition of logarithms this means that: