The Four Basic Rules

Idea

Four results follow straight from the definition. They are not new facts to memorise separately, and none of them needs a calculator.

Every one of them is the same sentence read back: $\log_a x$ is the power that turns $a$ into $x$ .

Rule

$\log_a 1 = 0$

$\log_a a = 1$

$\log_a a^n = n$

$a^{\log_a x} = x$

How it is used

1. Each rule is the definition read back

Logarithm form	What it asks	Index form
$\log_a 1 = 0$	which power gives $1$	$a^0 = 1$
$\log_a a = 1$	which power gives $a$	$a^1 = a$
$\log_a a^n = n$	which power gives $a^n$	$a^n = a^n$

The third rule contains the other two: put $n = 0$ and $n = 1$ .

2. The exponent walks straight out

Rewrite the argument as a power of the base, then read off the exponent.

\log_2 8 = \log_2 2^3 = 3,

\log_3 \frac{1}{9} = \log_3 3^{-2} = -2,

\log_5 \sqrt{5} = \log_5 5^{1/2} = \frac{1}{2}

Roots and fractions are not special cases. Write them as powers and the rule handles them.

3. The pair cancels

$a^{\log_a x} = x$ says the exponential and the logarithm undo each other on the spot.

3^{\log_3 7} = 7,

10^{\log_{10} 4} = 4

The bases must match. $2^{\log_3 7}$ simplifies to nothing.

4. Recognising a power of the base

This is the whole practical skill. Ask what power of the base produces the argument.

Exponential and logarithm

Try this. Compare bases 2 and 0.5: growth becomes decay. Move the input to 1; the logarithm is zero for either base.

BasePositive input x0.5

Base 3 · Blue: y = 3ˣ · Green: y = log_b(x) · Dashed: y = x. At x = 0.5, y = -0.631.

Argument	As a power	Value
$\log_2 32$	$2^5$	$5$
$\log_4 1$	$4^0$	$0$
$\log_7 7$	$7^1$	$1$
$\log_2 \dfrac{1}{16}$	$2^{-4}$	$-4$
$\log_3 \sqrt[3]{3}$	$3^{1/3}$	$\dfrac{1}{3}$

Worked example

Evaluate $\log_2 32 + \log_3 \dfrac{1}{27} - \log_5 \sqrt{5} + 4^{\log_4 6}$ .

Remember

$\log_a a^n = n$ , so write the argument as a power of the base $a^{\log_a x} = x$ when the two bases match $\sqrt{a} = a^{1/2}$ and $\dfrac{1}{a^n} = a^{-n}$

Take each term separately. First, write $32$ as a power of $2$ .

\log_2 32 = \log_2 2^5 = 5

A fraction is a negative power.

\log_3 \frac{1}{27} = \log_3 3^{-3} = -3

A square root is the power $\frac{1}{2}$ , and the last term is a matching pair.

\log_5 \sqrt{5} = \frac{1}{2},

4^{\log_4 6} = 6

Put the four values together.

5 + (-3) - \frac{1}{2} + 6 = \frac{15}{2}

Explore

Drag the probe along the green logarithm curve.

Two points on it are pinned by the first two rules. The curve crosses the $x$ -axis at $x = 1$ , because $\log_a 1 = 0$ whatever the base is. Change the base with the slider and watch that crossing refuse to move.

The second point does move. The height reaches $1$ directly above $x = a$ , because $\log_a a = 1$ , so sliding the base drags that point along with it. Read the base off the slider, then find the matching point on the curve.