## 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.
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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:

+

${\mathrm{log}}_{a}{\left(x\right)}^{n}=n{\mathrm{log}}_{a}x$

${\mathrm{log}}_{a}x=\frac{{\mathrm{log}}_{b}x}{{\mathrm{log}}_{b}a}$

${\mathrm{log}}_{a}a=$ 1

${\mathrm{log}}_{a}1= 0$

#### Examples

1. ${\mathrm{log}}_{2}5$ + ${\mathrm{log}}_{2}3$ = ${\mathrm{log}}_{2}\left(5\mathrm{x}3\right)$ = ${\mathrm{log}}_{2}15$

2. ${\mathrm{log}}_{3}10$ - ${\mathrm{log}}_{3}2$ = ${\mathrm{log}}_{3}\left(\frac{10}{2}\right)$ = ${\mathrm{log}}_{3}5$

3. ${\mathrm{log}}_{10}{5}^{3}$ = 3${\mathrm{log}}_{10}5$

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

= ${\mathrm{log}}_{10}6$ + ${\mathrm{log}}_{10}14$ - ${\mathrm{log}}_{10}7$ + ${\mathrm{log}}_{10}8$

= ${\mathrm{log}}_{10}\left(6×14×8/7\right)$

= ${\mathrm{log}}_{10}96$

5. ${\mathrm{log}}_{4}8$ = $\frac{{\mathrm{log}}_{2}8}{{\mathrm{log}}_{2}4}$

= $\frac{3}{2}$

= 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 loge or ln. For example, ln(7.389) = loge(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 $x$:

${\mathrm{log}}_{2}\left(3x-1\right)=3$

##### Solution

${\mathrm{log}}_{2}\left(3x-1\right)=3$ so that

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

$3x-1=8$ that is

$3x=8+1$ hence

$3x=9$ hence

$x=3$

#### Example 2

Solve the following for $x$:

${2\mathrm{log}}_{a}2x-{\mathrm{log}}_{a}\left(x-1\right)={\mathrm{log}}_{a}\left(4x-1\right)$

##### Solution

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

${\mathrm{log}}_{a}{\left(2x\right)}^{2}-{\mathrm{log}}_{a}\left(x-1\right)={\mathrm{log}}_{a}\left(4x-1\right)$ that is

${\mathrm{log}}_{a}4{x}^{2}-{\mathrm{log}}_{a}\left(x-1\right)={\mathrm{log}}_{a}\left(4x-1\right)$ that is

${\mathrm{log}}_{a}\frac{4{x}^{2}}{\left(x-1\right)}={\mathrm{log}}_{a}\left(4x-1\right)$ so that

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

$4{x}^{2}=\left(x-1\right)×\left(4x-1\right)$ that is

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

$5x=1$ hence

$x=0.2$

#### Example 3

Find $y$ in terms of $x$:

${\mathrm{log}}_{a}\left(x+1\right)+2{\mathrm{log}}_{a}y=3{\mathrm{log}}_{a}y-{\mathrm{log}}_{a}3x+{\mathrm{log}}_{a}x$

Solution

${\mathrm{log}}_{a}\left(x+1\right)+2{\mathrm{log}}_{a}y=3{\mathrm{log}}_{a}y-{\mathrm{log}}_{a}3x+{\mathrm{log}}_{a}x$ so that

${\mathrm{log}}_{a}\left(x+1\right)+{\mathrm{log}}_{a}3x-{\mathrm{log}}_{a}x=3{\mathrm{log}}_{a}y-2{\mathrm{log}}_{a}y$ that is

${\mathrm{log}}_{a}\left(x+1\right)+{\mathrm{log}}_{a}3x-{\mathrm{log}}_{a}x={\mathrm{log}}_{a}y$ so that

${\mathrm{log}}_{a}\frac{\left(x+1\right)×\left(3x\right)}{x}={\mathrm{log}}_{a}y$ so that

$\frac{\left(x+1\right)×\left(3x\right)}{x}=y$ that is

$y=\left(x+1\right)×3$ hence

$y=3x+3$