Logarithms - Maths & Stats - Powers and Logarithms - The Library at Leeds Beckett University

Learning Outcomes

If you work through this section you should be able to:

Explain the term 'logarithm'.
Understand the rules of logarithms and use them to simplify expressions.
Show how to change the base.
Solve exponential and logarithmic equations.

A power or exponent of a number states how many times to multiply the number by itself. The power is called the logarithm of a number to the base.

For example, 2⁵ = 32, then 5 = log₂⁡ 32, which is spoken as 5 is the logarithm of 32 to the base 2.

Examples

Exponential form	Logarithmic form
3⁴ = 81	4 = log₃ 81
6³ = 216	3 = log₆ 216
10⁵ = 100000	5 = log₁₀ 100000

Because logarithms are powers, the rules of logarithms closely follow the rules of indices. The rules of logarithms are given as below:

$\log_{a} (x y) = \log_{a} x$ + $\log_{a} y$

$\log_{a} (\frac{x}{y}) = \log_{a} x - \log_{a} y$

$\log_{a} {(x)}^{n} = n \log_{a} x$

$\log_{a} x = \frac{\log_{b} x}{\log_{b} a}$

$\log_{a} a =$ 1

$\log_{a} 1 = 0$

Examples

1. $\log_{2} 5$ + $\log_{2} 3$ = $\log_{2} (5 x 3)$ = $\log_{2} 15$

2. $\log_{3} 10$ - $\log_{3} 2$ = $\log_{3} (\frac{10}{2})$ = $\log_{3} 5$

3. $\log_{10} 5^{3}$ = 3 $\log_{10} 5$

4. $\log_{10} 6$ + $\log_{10} 14$ - $\log_{10} 7$ + 3 $\log_{10} 2$ = $\log_{10} 6$ + $\log_{10} 14$ - $\log_{10} 7$ + $\log_{10} 2^{3}$

= $\log_{10} 6$ + $\log_{10} 14$ - $\log_{10} 7$ + $\log_{10} 8$

= $\log_{10} (6 \times 14 \times 8 / 7)$

= $\log_{10} 96$

5. $\log_{4} 8$ = $\frac{\log_{2} 8}{\log_{2} 4}$

= $\frac{3}{2}$

= 1.5

When a logarithm is written without a base, it usually means that the base is 10. It is called common logarithm. For example, log⁡100 = log₁₀⁡100 = 2.

Another base that is often used is e (Euler's Number) which is about 2.71828. Logarithms to base e are called natural logarithms, which are denoted by log_e or ln. For example, ln(7.389) = log_e(7.389) ≈ 2.

Negative logarithms state how many to divide by the number.

For example, 2^-4 = $\frac{1}{2^{4}}$ = 0.0625, therefore log₂ 0.0625 = −4.

We will now look at how to use logarithms to solve logarithmic equations.

Example 1

Solve the following for $x$ :

$\log_{2} (3 x - 1) = 3$

Solution

$\log_{2} (3 x - 1) = 3$ so that

$3 x - 1 = 2^{3}$ hence

$3 x - 1 = 8$ that is

$3 x = 8 + 1$ hence

$3 x = 9$ hence

$x = 3$

Example 2

Solve the following for $x$ :

${2 \log}_{a} 2 x - \log_{a} (x - 1) = \log_{a} (4 x - 1)$

Solution

${2 \log}_{a} 2 x - \log_{a} (x - 1) = \log_{a} (4 x - 1)$ so that

$\log_{a} {(2 x)}^{2} - \log_{a} (x - 1) = \log_{a} (4 x - 1)$ that is

$\log_{a} 4 x^{2} - \log_{a} (x - 1) = \log_{a} (4 x - 1)$ that is

$\log_{a} \frac{4 x^{2}}{(x - 1)} = \log_{a} (4 x - 1)$ so that

$\frac{4 x^{2}}{(x - 1)} = 4 x - 1$ that is

$4 x^{2} = (x - 1) \times (4 x - 1)$ that is

$4 x^{2} = 4 x^{2} - 5 x + 1$ that is

$5 x = 1$ hence

$x = 0.2$

Example 3

Find $y$ in terms of $x$ :

$\log_{a} (x + 1) + 2 \log_{a} y = 3 \log_{a} y - \log_{a} 3 x + \log_{a} x$

Solution

$\log_{a} (x + 1) + 2 \log_{a} y = 3 \log_{a} y - \log_{a} 3 x + \log_{a} x$ so that

$\log_{a} (x + 1) + \log_{a} 3 x - \log_{a} x = 3 \log_{a} y - 2 \log_{a} y$ that is

$\log_{a} (x + 1) + \log_{a} 3 x - \log_{a} x = \log_{a} y$ so that

$\log_{a} \frac{(x + 1) \times (3 x)}{x} = \log_{a} y$ so that

$\frac{(x + 1) \times (3 x)}{x} = y$ that is

$y = (x + 1) \times 3$ hence

$y = 3 x + 3$

Activity

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Logarithms activity

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Skills for Learning: Maths & Stats - Powers and Logarithms

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Example 2

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Example 3

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