Laws of indices: Grade 12 Mathematics

By the end of this lesson you should be able to

Use the laws of indices to simplify expressions
Work with zero, negative and fractional indices
Solve simple equations where the unknown is an index

What is an index?

In the expression $a^n$ , the number $a$ is the base and $n$ is the index (also called the power or exponent). The index tells you how many times the base is multiplied by itself, so $2^4 = 2 \times 2 \times 2 \times 2 = 16$ .

Indices appear everywhere in algebra: in standard form, in formulae for area and volume, in growth and decay problems, and in equations. Learn the laws below well, because almost every later algebra topic uses them.

Key ideas

The laws of indices (for any base $a \neq 0$ )

Law	Rule	Example
Multiplying	$a^m \times a^n = a^{m+n}$	$2^3 \times 2^4 = 2^7$
Dividing	$a^m \div a^n = a^{m-n}$	$5^6 \div 5^2 = 5^4$
Power of a power	$(a^m)^n = a^{mn}$	$(3^2)^4 = 3^8$
Power of a product	$(ab)^n = a^n b^n$	$(2x)^3 = 8x^3$
Zero index	$a^0 = 1$	$7^0 = 1$
Negative index	$a^{-n} = \dfrac{1}{a^n}$	$2^{-3} = \dfrac{1}{8}$
Fractional index	$a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^m$	$8^{\frac{2}{3}} = 4$

Two points are worth remembering:

The laws for multiplying and dividing only work when the bases are the same. You can combine $x^3 \times x^5$ , but not $x^3 \times y^5$ .
A negative index does not make the number negative. It means “take the reciprocal”: $2^{-3}$ is $\frac{1}{8}$ , not $-8$ .

Solving equations with indices. If two powers with the same base are equal, their indices must be equal. In symbols, if $a^x = a^y$ then $x = y$ (for $a > 0$ and $a \neq 1$ ). To use this, write both sides with the same base.

Worked examples

Example 1. Simplify $\dfrac{x^5 \times x^3}{x^4}$ .

Add the indices on top, then subtract the index below:

\frac{x^{5+3}}{x^4} = \frac{x^8}{x^4} = x^{8-4} = x^4

Example 2. Evaluate $27^{\frac{2}{3}}$ .

The denominator of the fraction is the root and the numerator is the power. Take the root first because it keeps the numbers small:

27^{\frac{2}{3}} = \left(\sqrt[3]{27}\right)^2 = 3^2 = 9

Example 3. Evaluate $\left(\dfrac{2}{3}\right)^{-3}$ .

A negative index flips the fraction. Then apply the positive index to top and bottom:

\left(\frac{2}{3}\right)^{-3} = \left(\frac{3}{2}\right)^{3} = \frac{27}{8}

Example 4. Solve $2^{x+1} = 32$ .

Write $32$ as a power of $2$ : $32 = 2^5$ . Now both sides have base $2$ , so the indices are equal:

x + 1 = 5 \quad\Rightarrow\quad x = 4

Check: $2^{4+1} = 2^5 = 32$ . ✓

Example 5. Solve $9^x = 27$ .

Neither number is a power of the other, but both are powers of $3$ : $9 = 3^2$ and $27 = 3^3$ .

(3^2)^x = 3^3 \quad\Rightarrow\quad 3^{2x} = 3^3 \quad\Rightarrow\quad 2x = 3 \quad\Rightarrow\quad x = \frac{3}{2}

Example 6. Simplify $\dfrac{(2a^3b^{-2})^2}{4a^2b^{-1}}$ , leaving your answer with positive indices.

Square the top, handling the number, $a$ and $b$ separately:

\frac{4a^6b^{-4}}{4a^2b^{-1}}

The numbers cancel. Subtract the indices of $a$ and of $b$ :

a^{6-2}\, b^{-4-(-1)} = a^4 b^{-3} = \frac{a^4}{b^3}

Common mistakes

Combining different bases. $2^3 \times 4^2$ is not $8^5$ . Rewrite $4$ as $2^2$ first: $2^3 \times 2^4 = 2^7$ .

Treating $(a + b)^2$ as $a^2 + b^2$ . The power of a product rule works for multiplication only. $(a+b)^2 = a^2 + 2ab + b^2$ .

Mixing up $-3^2$ and $(-3)^2$ . Without brackets, only the $3$ is squared, so $-3^2 = -9$ . With brackets, $(-3)^2 = 9$ .

Multiplying the indices when you should add them. For $x^3 \times x^4$ the answer is $x^7$ , not $x^{12}$ . Multiply indices only for a power of a power, such as $(x^3)^4 = x^{12}$ .

Practice questions

Attempt every question on paper before you open the answers.

Simplify $y^7 \times y^{-3}$ .
Evaluate $16^{\frac{3}{4}}$ .
Evaluate $5^{-2}$ .
Simplify $\dfrac{6x^4y^3}{2x^2y}$ .
Solve $3^{x-2} = 81$ .
Solve $4^x = 8$ .
Evaluate $\left(\dfrac{1}{4}\right)^{-\frac{1}{2}}$ .
Simplify $(3p^2q^{-1})^3$ , leaving your answer with positive indices.

Show answers

$y^{7+(-3)} = y^4$
$\left(\sqrt[4]{16}\right)^3 = 2^3 = 8$
$\dfrac{1}{5^2} = \dfrac{1}{25}$
$3x^{4-2}y^{3-1} = 3x^2y^2$
$81 = 3^4$ , so $x - 2 = 4$ and $x = 6$
$4 = 2^2$ and $8 = 2^3$ , so $2^{2x} = 2^3$ , giving $2x = 3$ and $x = \dfrac{3}{2}$
$\left(\dfrac{1}{4}\right)^{-\frac{1}{2}} = 4^{\frac{1}{2}} = \sqrt{4} = 2$
$27p^6q^{-3} = \dfrac{27p^6}{q^3}$

Summary

Add indices when multiplying powers of the same base, subtract when dividing, and multiply for a power of a power.
$a^0 = 1$ , and $a^{-n}$ means $\dfrac{1}{a^n}$ .
In a fractional index, the denominator is the root and the numerator is the power.
To solve an equation with an unknown index, write both sides with the same base and equate the indices.

Now practise this topic

Real questions show you what still needs work. Try a past paper under timed conditions, or take a short quiz.

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