Introduction to Logarithms – A Level Maths Revision

A logarithm (usually abbreviated to ‘log’) is the inverse of an exponential function. It is a new topic at A Level, building on what you learnt at GCSE about powers and indices. It’s important to get a thorough introduction to logarithms – you’ll be building on this as you move onto logarithmic graphs and logarithmic equations.

If you want to see the introduction to logarithms below in PDF or presentable format, with a wider range of questions and answers, click here. For a multiple-choice quiz practicing the required prior knowledge, click here. If you think you’ve mastered logarithms, try these logarithm games or these exam-style questions.

As an example of an exponential:

103 = 1000

The same statement written as a logarithm is:

log10 1000 = 3

The small number next to ‘log’ is the base – this is the power we are working with. This example would be said as ‘log to base 10 of 1000 equals 3’. The left side of the statement above is saying ‘To what power do I need to raise 10, to get 1000?’, the answer to which is 3.

In more general terms:

If 𝑎𝑐 = 𝑏 then log𝑎 𝑏 = 𝑐

Note that a logarithm is only defined if the base (in this case, a) is a positive, real number not equal to 1.

Example Question 1

Rewrite the following using logarithms:

a. 52 = 25

This question is dealing with powers of 5, so the base is 5. What power of 5 is needed to make 25:

log5 25 = 2

b. 60 = 1

log6 1 = 0

Example Question 2

Rewrite the following using indices:

a. log9 81 = 2

Here, the base is 9. The statement is saying that 9 needs to be raised to the power of 2 to get 81:

92 = 81

b. $\boldsymbol{\log_{\frac{1}{2}}{\frac{1}{8}} = 2}$

$(\frac{1}{2})^3 = \frac{1}{8}$

In some cases, you can calculate logs mentally. For example, if you’re asked to find log2 32, you should be able to work that out by asking yourself what power of 2 gives 32. The answer is 5, as 25 = 32, therefore log2 32 = 5.

Example Question 3 (non-calculator)

Evaluate:

a. log3 81

34 = 81 therefore log3 81 = 4

b. log16 2

$16^{\frac{1}{4}}$ = 2, so log16 2 = $\frac{1}{4}$

However, solving logarithms mentally is not always possible. If you’re asked to find log2 17, you are unlikely to be able to find a solution mentally, because 17 is not an integer power of 2.

That’s where your calculator comes in. To find log2 17, use the button, being careful to put 2 and 17 in the correct places. This will give 4.09 (2d.p).

Example Question 4 (calculator)

Evaluate the following. Give you answers correct to 3s.f.:

a. log3 21

log3 21 = 2.77

b. $\boldsymbol{ \log_{\frac{1}{3}}{8}}$

$\log_{\frac{1}{3}}{8} = -1.89$

There are three log laws which you’ll need to know and apply; they’re similar to the laws of indices. The laws are:

\begin{aligned} &\log_a{b} + \log_a{c} = \log_a{bc} \\ &\log_a{b} - \log_a{c} = \log_a{\frac{b}{c}} \\ &\log_a{b^c}=c\log_a{b} \end{aligned}\begin{aligned} &\log_a{b} + \log_a{c} = \log_a{bc} \\ &\log_a{b} - \log_a{c} = \log_a{\frac{b}{c}} \\ &\log_a{b^c}=c\log_a{b} \end{aligned}

Test them out. You could let a = 10, b = 1000 and c = 100 in the first law if you want numbers that you can work with mentally, or you could use any positive numbers you like if you have a calculator to hand.

Example Question 5

Write each question as a single logarithm

a. log3 9 + log3 81

= log3 (9 x 81)

= log3 729

b. loga 15 – loga 90

= loga (15 ÷ 90)

= loga $\frac{15}{90}$

c. 2log3 9 + log3 11

= log3 (92) + log3 11

= log3 (81 x 11)

= log3 891

Summary

log𝑎 𝑏 = 𝑐 is equivalent to 𝑎𝑐 = 𝑏
Where 𝑎 is a positive, real number not equal to 1.

log𝑎 𝑏 + log𝑎 𝑐 = log𝑎 𝑏𝑐
log𝑎 𝑏 - log𝑎 𝑐 = log𝑎 $\frac{b}{c}$$\frac{b}{c}$
log𝑎 (𝑏𝑐) = clog𝑎 𝑏

To evaluate a logarithm on your calculator, use the  key.

Introduction to Logarithms – Practice Questions

Non-Calculator

1. Complete the table, using equivalent statements:

2. For each question, find the value of 𝑥:

a. log𝑥 9 = 2

b. log6 𝑥 = -2

c. log11 1331 = 𝑥

d. log20 400 = 𝑥

e. log4𝑥 16 = 4

f. log2 8𝑥 = 6

3. Express each as a single logarithm:

a. log3 8 + log3 2

b. 2log2 𝑎 + log2 𝑏

c. log𝑦 8 – 2log𝑦 4

d. log2 2 – log2 0.5 – log2 8

e. 4log𝑝 𝑚 – log𝑝 2𝑚

f. log10 3 + 0.5log10 9 – log10 19

4. Given that log𝑡 𝑎 = 𝑥 and log𝑡 𝑏 = 𝑦, express log𝑡 (𝑎2𝑏3) in terms of 𝑥 and 𝑦.

5. 𝑎 = log𝑐 2 and 𝑏 = log𝑐 3. Express log𝑐 48 in terms of 𝑎 and 𝑏.

Calculator

a. log7 111

b. log0.3 5

c. log6 9

d. log0.1 0.38

1.

2.

a. log𝑥 9 = 2

𝑥2 = 9

𝑥 = 3

b. log6 𝑥 = -2

6-2 = 𝑥

𝑥 = $\frac{1}{36}$

c. log11 1331 = 𝑥

11𝑥 = 1331

𝑥 = 3

d. log20 400 = 𝑥

𝑥 = 2

e. log4𝑥 16 = 4

𝑥 = 0.5

f. log2 8𝑥 = 6

𝑥 = 8

3. Express each as a single logarithm:

a. log3 8 + log3 2

log3 16

b. 2log2 𝑎 + log2 𝑏

log2 𝑎2𝑏

c. log𝑦 8 – 2log𝑦 4

= log𝑦 8 – logy (42)

= log𝑦 $\frac{8}{16}$

= log𝑦 $\frac{1}{2}$

d. log2 2 – log2 0.5 – log2 8

= log2 $\frac{2}{0.5}$ – log2 8

= log2 4 – log2 8

= log2 $\frac{1}{2}$

e. 4log𝑝 𝑚 – log𝑝 2𝑚

= log𝑝 (𝑚4) – logp 2𝑚

= log𝑝 $(\frac{m^4}{2m})$

= log𝑝 $(\frac{m^4}{2m})$

f. log10 3 + 0.5log10 9 – log10 19

= log10 3 + log10 (90.5) – log10 19

= log10 3 + log10 3 – log10 19

= log10 9 – log10 19

= log10 $\frac{9}{19}$

4. Given that log𝑡 𝑎 = 𝑥 and log𝑡 𝑏 = 𝑦, express log𝑡 (𝑎2𝑏3) in terms of 𝑥 and 𝑦.

log𝑡 (𝑎2)(b3) = log𝑡 𝑎2 + log𝑡 𝑏3

= 2log𝑡 𝑎 + 3log𝑡 𝑏

= 2𝑥 + 3𝑦

5. 𝑎 = log𝑐 2 and 𝑏 = log𝑐 3. Express log𝑐 48 in terms of 𝑎 and 𝑏.

log𝑐 48 = log𝑐 (2 x 2 x 2 x 2 x 3)

= log𝑐 2 + log𝑐 2 + log𝑐 2 + log𝑐 2 + log𝑐 2 + log𝑐 3

= 4𝑎 + 𝑏