The laws of indices · Indices & standard form

In $2^5$ , the 2 is the base and the 5 is the index (or power): $2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32$ . Every law in this lesson comes from writing the powers out as repeated multiplication.

Try it

Why the laws of indices workChange m and n

aaa

aaaaa

a³ × a² = a⁵the law

m = 3n = 2

a³ is 3 a's multiplied together and a² is 2 more. Together that's 5 a's: add the indices. (This only works when the base is the same.)

Try each button. In aᵐ ÷ aⁿ, make $n$ bigger than $m$ . The leftover a’s end up on the bottom of the fraction: that is what a negative index means. Zero and negative shows the halving pattern carrying on below $2^1$ .

The laws

Law	Example
$a^m \times a^n = a^{m + n}$	$2^3 \times 2^4 = 2^7$
$a^m \div a^n = a^{m - n}$	$5^6 \div 5^2 = 5^4$
$(a^m)^n = a^{mn}$	$(3^2)^4 = 3^8$
$a^0 = 1$	$7^0 = 1$
$a^{-n} = \dfrac{1}{a^n}$	$2^{-3} = \dfrac18$
$a^{\frac1n} = \sqrt[n]{a}$	$8^{\frac13} = 2$
$a^{\frac mn} = \left(\sqrt[n]{a}\right)^m$	$8^{\frac23} = 2^2 = 4$

MultiplyingCount the a's: add the indices

DividingCancel the a's: subtract the indices

A power of a powerGroups of a's: multiply the indices

Zero and negativeKeep halving: 2⁰ = 1, 2⁻¹ = ½

Fractional indices: root first, then power

For $a^{\frac mn}$ , take the $n$ th root first, while the number is still small. Then raise it to the power $m$ :

\begin{aligned} 16^{\frac34} &= \left(\sqrt[4]{16}\right)^3 &&\text{root first}\\ &= 2^3 &&\sqrt[4]{16} = 2\\ &= 8 \end{aligned}

Root first, then powerThe bottom of the fraction is the root; the top is the power

Negative indices: turn the fraction over

A negative index means “one over”, so for a fraction it turns the fraction upside down:

\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^{n}

More: evaluating powers

Decimals: change to a fraction first

Worked example · WAEC 2019

WAEC 2019 · Paper 1 · Q2

Evaluate $(0.064)^{-\frac13}$ .

Write the decimal as a fraction
$0.064 = \frac{64}{1000}$ .
Think first. $0.064 = \frac{?}{1000}$
Deal with the minus sign
It turns it over: $\left(\frac{64}{1000}\right)^{-\frac13} = \left(\frac{1000}{64}\right)^{\frac13}$ .
Think first. What does the negative index do to the fraction?
Take the cube root
$\sqrt[3]{1000} = 10$ and $\sqrt[3]{64} = 4$ , so the answer is $\frac{10}{4} = 2.5$ .

Simplifying with letters and numbers

Write every number as a power of the same prime, then use the laws. A root of a term with letters works the same way, one piece at a time:

take the root of each piece: ${\sqrt[3]{125x^6} = 125^{\frac13} \times (x^6)^{\frac13}}$ ;
${125^{\frac13} = 5}$ , and ${6 \times \frac13 = 2}$ , so the answer is ${5x^2}$ .

More: simplifying with letters

When powers are added or taken away

The laws don’t apply to a sum such as $2^{n + 3} + 2^n$ . Take out the smallest power as a common factor instead:

\begin{aligned} 2^{n + 3} + 2^n &= 2^n(2^3 + 1) &&2^{n + 3} = 2^n \times 2^3\\ &= 9 \times 2^n \end{aligned}

Then the $2^n$ usually cancels.

More: sums of powers

Your turn

WAEC 2025 · Paper 1 · Q4 ✱✱

Indices & standard form

Simplify $0.027^{-\frac13}$ .

Worked solution (try it first)

$0.027 = \frac{27}{1000}$ , and its cube root is $\frac{3}{10} = 0.3$ .
The negative index means the reciprocal: $\frac{1}{0.3} = \frac{10}{3}$ .
As a mixed number, $\frac{10}{3} = 3\frac13$ , option B.

Report a problem with this question