Let a and N be positive real numbers and let N = a^{n}. Then n is called the logarithm of N to the base a. We write this as:
n = log_{a}N |
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 = 3^{3}, 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
Let a, M, N be positive real numbers and k be any number. Then the following important rules apply to logarithms:
Rule 1 | log_{a}(MN)^{ } | = | log_{a}M + log_{a}N^{ } |
---|---|---|---|
Rule 2 | log_{a}(MN)^{ } | = | log_{a}M^{ } − log_{a}N^{ } |
Rule 3 | log_{a}(m^{k}) | = | k log_{a}M^{ } |
Rule 4 | log_{a}a^{ } | = | 1 |
Rule 5 | log_{a}1^{ } | = | 0 |
Proof that log_{a} MN = log_{a} M + log_{a} N:
Let m = log_{a} M and n = log_{a} N, so, by definition, M = a^{m} and N = a^{n}. Then:
- where we have used the appropriate rule for exponents. From this, using the definition of a logarithm, we have:
But m + n = log_{a} M + log_{a} N, and the above equation may be written:
- which is what we wanted to prove.
Example 2:
If x = log_{6} 36, then 6^{x} = 36 = 6^{2}. Thus, log_{6} 4 + log_{6} 9 = 2.
Now 20 × 1 ⁄ 4 = 5, so log_{5} 20 + log_{4} (1 ⁄ 4) = log_{5} 5 = 1
Quiz 1: To which of the following numbers does the expression log_{3}15 + log_{3}0.6 simplify?
Solution:
Using Rule 1 we have: log_{3} 15 + log_{3} 0.6 = log_{3} (15 × 0.6) = log_{3} 9
But, 9 = 3^{2}, so log_{3} 15 + log_{3}0.6 = log_{3}3^{2} = 2.
Proof that log_{a}MN = log_{a} M − log_{a} N:
As before, let m = log_{a} M and n = log_{a} N. Then, M = a^{m} and N = a^{n}. 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 = log_{2}8, then 2^{x} = 8 = 2^{3} , so x = 3.
Since 3 ⁄ 5 = 0.6, then log_{3} 0.6 = log_{3} (35) = log_{3} 3 − log_{3} 5.
Now log_{3} 3 = 1, so that log_{3} 0.6 = 1 − 1.465 = −0.465
Quiz 2: To which of the following numbers does the expression log_{2}12 − log_{2}(3 ⁄ 4) simplify?
Solution:
Using Rule 2 we have: log_{2}12 − log_{2} (3 ⁄ 4) = log_{2} (12 ÷ 3)
Now we have 12 ÷ (3 ⁄ 4) = 12 × (4 ⁄ 3) = 12 × 43 = 16.
Thus log_{2} 12 − log_{2} (3 ⁄ 4) = log_{2} 16 = log_{2}2^{4}.
If x = log_{2}2^{4}, then 2^{x} = 2^{4} , so x = 4.
Proof that log_{a} (M^{k}) = k log_{a} M:
Let m = log_{a} M, so M = a^{m}. Then:
- where we have used the appropriate rule for indices. From this we have, by the definition of a logarithm:
But m = log_{a} M, so the last equation may be written:
- which is the result we wanted.
Example 4:
We have 10000 = 10^{4}, so 1 ⁄ 10000 = 1 ⁄ 10^{4} = 10^{−4}.
Thus, log_{10} (1 ⁄ 10000) = log_{10} (10^{−4}) = −4 log_{10} 10 = −4, where we have used Rule 4 to write log_{10} 10 = 1.
We have 6 = √36 = 36^{1 ⁄ 2} .
Thus, log_{36} 6 = log_{36} (36^{1 ⁄ 2}) = 12 log_{36} 36 = 12.
Quiz 3: If log_{3} 5 = 1.465, which of the following numbers is log_{3} 0.04?
Solution:
Note that: 0.04 = 4 ⁄ 100 = 1 ⁄ 25 = 1 ⁄ 52 = 5^{−2}.
Thus, log_{3} 0.04 = log_{3} (5^{−2}) = −2 log_{3} 5.
Since log_{3} 5 = 1.465, we have log_{3} 0.05 = −2 × 1.465 = −2.930
In this section we look at some applications of the rules of logarithms.
Example 5:
= log_{10} (10^{3}) = 3 log_{10} 10 = 3.
= log_{10} 100 = log_{10} (10^{2}) = 2 log_{10} 10 = 2.
= log_{a} (4^{3} × 14) − log_{a} (2^{4}) = log_{a} (4^{2}) − log_{a} (2^{4})
= log_{a} 16 − log_{a} 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 log_{3} 2 = log_{3} (2^{3}) = log_{3} 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 log_{a} 6 − (log_{a} 4 + 2 log_{a} 3) | = | log_{a} (6^{2}) − log_{a} (4 × 3^{2}) |
= | log_{a} (6^{2}4 × 3^{2}) | |
= | log_{a} 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 log_{3} 6 − (2 log_{3} 4 + log_{3} 18) | = | log_{3} (6^{5}) − log_{3} (4^{2} × 18) |
= | log_{3} (6^{5}4^{2} × 18) | |
= | log_{3} (2^{5} × 3^{5}4^{2} × 2 × 9) | |
= | log_{3}(3^{3}) | |
= | 3 log_{3} 3 = 3 |
- since log_{3} 3 = 1
Solution:
The first thing we note is that √3 can be written as 3^{½}. We first simplify some of the terms. They are:
and
Putting all of this together:
3 log_{4} ( √3) − 12 log_{4} 2 + 3 log_{4} 2 − log_{4} 6 | = | 32 log_{4} 3 − 12 log_{4}3 + 3 log_{4} 2 − (log_{4} 2 + log_{4} 3) |
= | (32 − 12 − 1) log_{4} 3 + (3 − 1) log_{4} 2 | |
= | 2 log_{3} 2 = log_{4} (2^{2}) = log_{4} 4 = 1 |
Click on questions to reveal solutions
In each of the following, find x
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:
log_{a}c = log_{a}b × log_{b}c |
Proof of the above rule:
Let x = log_{a} b and y = log_{b} c. Then, by the definition of logarithms:
This means that:
- with the last equality following from the laws of indices. Since c = a^{xy} , by the definition of logarithms this means that: