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:

(a)Since 16 = 24, then 4 = log216
(b)Since 81 = 34, then 4 = log381
(c)Since 3 = 9, then 1 ⁄ 2 = log93
(d)Since 3−1 = 1 ⁄ 3, then −1 = log3 (1 ⁄ 3)

Exercise 1: Use the definition of logarithms given above to determine the value of x in each of the following:

(a)x = log3 27

Solution:

By the definition of a logarithm, we have:

3x = 27

But 27 = 33, so we have:

3x = 27 = 33

So:

x = 3
(b)x = log5 125

Solution:

By the definition of a logarithm, we have:

25x = 5

Now:

5 = 25 = 25½

So:

25x = 5 = 25½

From this we see that x = ½

(c)x = log2 (1 ⁄ 4)

Solution: By the definition of a logarithm, we have:

2x = 1 ⁄ 4 = 1 ⁄ (22) = 2−2

Thus x = −2

(d)2 = logx(16)

Solution: By the definition of a logarithm, we have:

x2 = 16 = 42

Thus x = 4

(e)3 = log2x

Solution: By the definition of a logarithm, we have:

23 = x

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 Rule 2 Rule 3 loga(MN) = logaM + logaN loga(M/N) = logaM  − logaN loga(mk) = k logaM logaa = 1 loga1 = 0

3. Logarithm of a Product

Proof that loga MN = loga M + loga N:

Let m = logaM and n = logaN, so, by definition, M = am and N = an. Then:

MN = am × an = am + n

- where we have used the appropriate rule for exponents. From this, using the definition of a logarithm, we have:

m + n = loga (MN)

But m + n = loga M + loga N, and the above equation may be written:

loga M + loga N = loga (MN)

- which is what we wanted to prove.

Example 2:

(a)log64 + log69 = log6(4 × 9) = log636

If x = log6 36, then 6x = 36 = 62. Thus, log6 4 + log6 9 = 2.

(b)log5 20 + log4 1 ⁄ 4 = log5 (20 × 1 ⁄ 4)

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?

(c)2Correct - well done!

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 logaM/N = loga M − loga N:

As before, let m = logaM and n = logaN. Then, M = am and N = an. Now we have:

M/N = am/an = am − n

- where we have used the appropriate rule for indices. By the definition of a logarithm, we have:

mn = loga(M/N)

From this we are able to deduce that:

logaM − logaN = mn = loga(M/N)

Example 3:

(a)log2 40 − log2 5 = log2 (40 ⁄ 5) = log2 8

If x = log28, then 2x = 8 = 23 , so x = 3.

(b)If log35 = 1.465 then we can find log3 0.6.

Since 3 ⁄ 5 = 0.6, then log3 0.6 = log3(3/5) = 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?

(d)4Correct - well done!

Solution:

Using Rule 2 we have: log212 − log2 (3 ⁄ 4) = log2 (12 ÷ 3)

Now we have 12 ÷ (3 ⁄ 4) = 12 × (4 ⁄ 3) = 12 × 4/3 = 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 = logaM, so M = am. Then:

Mk = (am)k = amk = akm

- where we have used the appropriate rule for indices. From this we have, by the definition of a logarithm:

km = loga(Mk)

But m = logaM, so the last equation may be written:

k logaM = km = loga (Mk)

- which is the result we wanted.

Example 4:

(a)Find log10 (1 ⁄ 10000).

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.

(b)Find log36 6.

We have 6 = 36 = 361 ⁄ 2 .

Thus, log36 6 = log36 (361 ⁄ 2) = 1/2 log36 36 = 1/2.

Quiz 3: If log3 5 = 1.465, which of the following numbers is log3 0.04?

(a)−2.930Correct - well done!

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:

(a)log4 1 = 0.
(b)log10 10 = 1.
(c)log10 125 + log10 8 = log10 (125 × 8) = log10 1000

= log10 (103) = 3 log10 10 = 3.

(d)2 log10 5 + log10 4 = log10 (52) + log10 4 = log10 (25 × 4)

= log10 100 = log10 (102) = 2 log10 10 = 2.

(e)3 loga 4 + loga (1 ⁄ 4) − 4 loga 2 = loga (43) + loga (1 ⁄ 4) − loga (24)

= loga (43 × 1/4) − loga (24) = loga (42) − loga (24)

= loga 16 − loga 16 = 0.

Exercise 2: Use the rules of logarithms to simplify each of the following:

(a)3 log3 2 − log3 4 + log3(1/2)

Solution:

First of all, by Rule 3, we have 3 log3 2 = log3 (23) = log3 8. Thus, the expression becomes:

log3 8 − log3 4 + log3 (1/2) = [log3 8 + log3 (1/2)] − log3 4.

Using Rule 1, the term in the square brackets becomes:

log3(8 × 1/2) = log3 4

The expression then simplifies to:

log3 4 − log3 4 = 0
(b)3 log10 5 + 5 log10 2 − log10 4

Solution:

First we use Rule 3 to rewrite the first two terms:

3 log10 5 = log10 (53)

and:

5 log10 2 = log10 (25)

Thus:

3 log10 5 + 5 log10 2 = log10 (53) + log10 (25) = log10 (53 × 25)

- where we have used Rule 1 to obtain the right hand side. Thus:

3 log10 5 + 5 log10 2 − log10 4 = log10 (53 × 25) − log10 4

and, using Rule 2, this simplifies to:

log10(53 × 25/4) = log10 (103) = 3 log10 10 = 3
(c)2 loga 6 − (loga 4 + 2 loga 3)

Solution:

Dealing first with the expression in brackets, we have:

loga 4 + 2 loga 3 = loga 4 + loga (32) = loga (4 × 32)

- where we have used, in succession, Rules 3 and 2. Now:

2 loga 6 = loga (62)

so that finally, we have:

 2 loga 6 − (loga 4 + 2 loga 3) = loga (62) − loga (4 × 32) = loga (62/4 × 32) = loga 1 = 0
(d)5 log3 6 − (2 log3 4 + log3 18)

Solution:

Dealing first with the expression in brackets, we have:

2 log3 4 + log3 18 = log3 (42) + log3 18 = log3 (42 × 18)

- 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 (65/42 × 18) = log3 (25 × 35/42 × 2 × 9) = log3(33) = 3 log3 3 = 3

- since log3 3 = 1

(e)3 log4 ( 3) − 1/2 log4 2 + 3 log4 2 − log4 6

Solution:

The first thing we note is that 3 can be written as 3½. We first simplify some of the terms. They are:

3 log4 (3) = 3 log4 (3½) = 3/2 log4 3

and

log4 6 = log4 (2 × 3) = log4 2 + log4 3

Putting all of this together:

 3 log4 ( √3) − 1/2 log4 2 + 3 log4 2 − log4 6 = 3/2 log4 3 − 1/2 log43 + 3 log4 2 − (log4 2 + log4 3) = (3/2 − 1/2 − 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

1.logx 1024 = 2
(a)23
(b)24
(c)22
(d)25
2.x = (loga27 − loga8 − loga125)/ (loga 6 − loga 20)
(a)1
(b)3
(c)3 ⁄ 2
(d)−2 ⁄ 3
3.logc (10 + x) − logcx = logc 5
(a)2.5
(b)4.5
(c)5.5
(d)7.5

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 = logab   and   y = logbc. Then, by the definition of logarithms:

ax = b and by = c

This means that:

c = by = (ax)y = axy

- with the last equality following from the laws of indices. Since c = axy , by the definition of logarithms this means that:

logac = xy = logab × logbc