Logarithms and their laws · Logarithms

A logarithm answers the question “what power?”. Since $2^5 = 32$ , the logarithm of 32 to base 2 is 5:

\log_a x = n \quad\text{means}\quad a^n = x

Two ways to say one factThe log is the power

So $\log_{10} 1000 = 3$ , $\log_3 81 = 4$ and $\log_5 \frac{1}{25} = -2$ . When no base is written, it is 10. You need the laws of indices: every law of logarithms is an index law in disguise.

More: what a logarithm means

The laws

A slide rule: logs turn × into +Change a and b

0.301log 20.4771+ log 30.7782= log 662 × 3

a = 2b = 3

On these scales each number sits at a distance proportional to its logarithm. Slide the lower scale so its 1 is under 2: that's a length of log 2. Then go along the lower scale to 3, a further log 3. Above it you read 6, because log 2 + log 3 = log (2 × 3).

On a slide rule every number is placed at a distance equal to its logarithm. Adding two lengths lands you on the product: that’s the first law.

Law	Because
$\log_a(xy) = \log_a x + \log_a y$	multiplying powers adds indices
$\log_a\left(\dfrac{x}{y}\right) = \log_a x - \log_a y$	dividing powers subtracts indices
$\log_a x^n = n\log_a x$	a power of a power multiplies indices
$\log_a a = 1$ , $\log_a 1 = 0$	$a^1 = a$ , $a^0 = 1$

Logs turn × into +On a log scale, lengths add when numbers multiply

Evaluating without tables

Bring everything into one logarithm: numbers in front become powers, pluses become multiplication, minuses become division. Then simplify the number inside.

Worked example · NECO 2023

NECO 2023 · Paper 2 · Q8 (a)

Evaluate without using tables $5\log2 + \log40 - \log12.8$ .

Move the 5 inside
$5\log 2 = \log 2^5 = \log 32$ .
Think first. What does $5\log 2$ become?
Combine into one log
$\log 32 + \log 40 - \log 12.8 = \log\frac{32 \times 40}{12.8}$
Simplify
$\frac{1280}{12.8} = 100$ , and $\log 100 = 2$ (since $10^2 = 100$ ).
Think first. What is $32 \times 40 \div 12.8$ ?

More: evaluating without tables

More: using given log values

Awkward bases

To evaluate a log such as $\log_8 4$ , where the answer isn’t a whole number, call it $n$ and change to index form. Then write both sides as powers of the same base, as in equations with indices.

Call it nIndex form, then the same base on both sides

Two more facts help here. Because a log is a power, $a^{\log_a x} = x$ : for example $3^{\log_3 7} = 7$ . And to change the base, for any base $b$ :

\log_a x = \frac{\log_b x}{\log_b a}

With $b = 10$ , this is how a calculator finds $\log_8 4$ : it works out $\log 4 \div \log 8 = 0.6021 \div 0.9031 = 0.667$ .

More: awkward bases

Equations with logarithms

Two methods, often together:

Combine the logs into a single log on each side, then drop the logs: $\log_a P = \log_a Q$ means $P = Q$ .
Change to index form: $\log_a P = n$ means $P = a^n$ . A plain number like 3 can be written as $\log_a a^3$ .

Worked example · WAEC 2017

WAEC 2017 · Paper 2 · Q7 (a)

If $\log_6 y + 2\log_6 x = 3$ , express $y$ in terms of $x$ .

Combine the left side
$2\log_6 x = \log_6 x^2$ , so $\log_6 y + \log_6 x^2 = \log_6(yx^2)$ .
Think first. What does $2\log_6 x$ become?
Change to index form
$\log_6(yx^2) = 3$ means $yx^2 = 6^3 = 216$ .
Make y the subject
$y = \frac{216}{x^2}$

More: relations without logs

More: equations with logarithms

Your turn

WAEC 2014 · Paper 2 · Q4 (a)

(a)
Without using tables or a calculator, evaluate $\log_{10}\left(\frac{75}{10}\right) - 2\log_{10}\left(\frac59\right) + \log_{10}\left(\frac{100}{243}\right)$ .

Worked solution (try it first)

(a)

Move the 2 inside as a power: $2\log_{10}\frac59 = \log_{10}\frac{25}{81}$ .
Then combine: $\log_{10}\left(\frac{75}{10} \times \frac{100}{243} \div \frac{25}{81}\right)$ .
Simplify: $\frac{75}{10} \times \frac{100}{243} \times \frac{81}{25} = \frac{75 \times 100 \times 81}{10 \times 243 \times 25}$
$= 10$ .
So the answer is $\log_{10} 10 = 1$ .

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