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 𝑎^𝑐 = 𝑏 then log_𝑎 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:

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. log_a 15 – log_a 90

= log_a (15 ÷ 90)

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

c. 2log₃ 9 + log₃ 11

= log₃ (9²) + log₃ 11

= log₃ (81 x 11)

= log₃ 891

Summary

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

log_𝑎 𝑏 + log_𝑎 𝑐 = log_𝑎 𝑏𝑐
log_𝑎 𝑏 - log_𝑎 𝑐 = log_𝑎  
log_𝑎 (𝑏𝑐) = clog_𝑎 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. log_𝑥 9 = 2

b. log₆ 𝑥 = -2

c. log₁₁ 1331 = 𝑥

d. log₂₀ 400 = 𝑥

e. log_4𝑥 16 = 4

f. log₂ 8𝑥 = 6

3. Express each as a single logarithm:

a. log₃ 8 + log₃ 2

b. 2log₂ 𝑎 + log₂ 𝑏

c. log_𝑦 8 – 2log_𝑦 4

d. log₂ 2 – log₂ 0.5 – log₂ 8

e. 4log𝑝 𝑚 – log𝑝 2𝑚

f. log₁₀3 + 0.5log₁₀ 9 – log₁₀ 19

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

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

Calculator

6. Giving your answers correct to 2 s.f., evaluate:

a. log₇ 111

b. log_0.3 5

c. log₆9

d. log_0.1 0.38

a. log_𝑥 9 = 2

𝑥² = 9

𝑥 = 3

b. log₆ 𝑥 = -2

6^-2 = 𝑥

𝑥 = $\frac{1}{36}$

c. log₁₁ 1331 = 𝑥

11^𝑥 = 1331

𝑥 = 3

d. log₂₀ 400 = 𝑥

𝑥 = 2

e. log_4𝑥 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. log_𝑦 8 – 2log_𝑦 4

= log_𝑦 8 – logy (42)

= log_𝑦 $\frac{8}{16}$

= log_𝑦 $\frac{1}{2}$

d. log₂ 2 – log₂ 0.5 – log₂ 8

= log₂ $\frac{2}{0.5}$ – log₂ 8

= log₂ 4 – log₂ 8

= log₂ $\frac{1}{2}$

e. 4log_𝑝 𝑚 – log_𝑝 2𝑚

= log_𝑝 (𝑚⁴) – logp 2𝑚

= log_𝑝 $(\frac{m^4}{2m})$

f. log₁₀3 + 0.5log₁₀ 9 – log₁₀ 19

= log₁₀ 3 + log₁₀ (9^0.5) – log₁₀ 19

= log₁₀ 3 + log₁₀ 3 – log₁₀ 19

= log₁₀ 9 – log₁₀ 19

= log₁₀ $\frac{9}{19}$

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

log_𝑡 (𝑎²)(b³) = log_𝑡 𝑎² + log_𝑡 𝑏³

= 2log_𝑡 𝑎 + 3log_𝑡 𝑏

= 2𝑥 + 3𝑦

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

log_𝑐 48 = log_𝑐 (2 x 2 x 2 x 2 x 3)

= log_𝑐 2 + log_𝑐 2 + log_𝑐 2 + log_𝑐 2 + log_𝑐 2 + log_𝑐 3

= 4𝑎 + 𝑏

6. Giving your answers correct to 2 s.f., evaluate:

a. log₇ 111 = 2.4

b. log_0.3 5 = -1.3

c. log₆9 = 1.2

d. log_0.1 0.38 = 0.42

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