# Indices and Index Laws – A Level Maths Revision

The basic rules for working with indices at A level are the same as at GCSE; the challenge is, you’re expected to be able to recognise and apply those rules as quickly and fluently as you could apply basic addition and subtraction in a GCSE question.

If you’re not quite there yet or you’d appreciate some guidance to help you brush up on the details, this post should be ideal for you. You’ll find a series of worked examples surrounded by supplementary information to help your end-game comprehension of Indices at A Level.

If you want a little more practice with GCSE level index laws question, have a look at this preparation for A level resource and these prior-knowledge multiple-choice questions. Or, if you’d like the explanation below in PDF or powerpoint format, along with questions and solutions, click here.

## Indices

Indices are a way of showing that we are taking a constant, variable or expression and raising it to a power, finding its root or taking its reciprocal. The index can take any value but, for now, we will be sticking to rational indices.

A positive, integer index is an instruction to multiply the base number by itself, where the index gives the number of bases to multiply. For example:

$a^5 = a \times a \times a \times a \times a$

$(x + 4)^3 = (x+4)(x+4)(x+4)$

A negative index represents a reciprocal, for example:

$5^{-1} =$ $\frac{1}{5}$

$(3x + 2)^{-1} =$ $\frac{1}{3x + 2}$

$2^{-3} =$ $\frac{1}{2^3} = \frac{1}{8}$

Or, in general terms:

$\boldsymbol{a^{-n} = \frac{1}{a^n}}$

A fraction index, with a numerator of 1, represents a root, for example:

$27^{\frac{1}{3}} = \sqrt[3]{27} = 3$

$(4x^2)^{\frac{1}{2}} = \sqrt{(4x^2)} = 2x$

A fraction index where the numerator is not 1 represents both a root and a power:

$8^{\frac{2}{3}} = \sqrt[3]{8^2} = \sqrt[3]{64} = 4$

Or, in more general terms:

$\boldsymbol{a^{\frac{b}{c}} = \sqrt[c]{a^b}}$

These can be combined in any form, so a power of $-\frac{3}{2}$ would represent taking the reciprocal (because it’s negative), cubing (because of the numerator of 3) and then square rooting (because of the denominator of 2). While the order in which you perform these operations may affect the difficulty of the calculation, it will not affect the result.

### Example Question 1

Write the following expression in the form $\boldsymbol{ (x - a)^b}$, where $\boldsymbol{x \neq a}$ and $\boldsymbol{b}$ is any rational number:
$\boldsymbol{\frac{1}{\sqrt[3]{x^2\, -\, 6x\, +\, 9}}}$

Start by factorising the expression on the denominator:

$x^2 - 6x + 9 = (x-3)^2$

This means that:

$\frac{1}{\sqrt[3]{x^2 \,- \,6x \,+ \,9}} = \frac{1}{\sqrt[3]{(x\,-\,3)^2}}$

We know that $a^{\frac{b}{c}} = \sqrt[c]{a^b}$, therefore we can re-write the expression as:

$\frac{1}{\sqrt[3]{(x\,-\,3)^2}} = \frac{1}{(x\,-\,3)^{\frac{2}{3}}}$

Finally, $a^{-n} = \frac{1}{a^n}$, therefore:

$\frac{1}{(x\,-\,3)^{\frac{2}{3}}} = (x-3)^{-\frac{2}{3}}$

## Index Laws

The index laws simplify the process of simplifying indices.

The multiplication law:

$a^2 \times a^3 = \underline{a \times a} \times \underline{a \times a \times a} = a^5$

Or, in more general terms:

$\boldsymbol{a^x \times a^y = a^{x\,+\,y} }$

The division law:

$\frac{a^5}{a^3} = \frac{a \, \times \, a \, \times \, a}{a \, \times \, a \, \times \, a} \times a \times a = 1 \times a \times a = a^2$

Or, in more general terms:

$\boldsymbol{\frac{a^x}{a^y} = a^{x\,-\,y} }$

The power law:

$(a^3)^2 = \underline{a \times a \times a} \times \underline{a \times a \times a} = a^6$

Or, in more general terms:

$\boldsymbol{(a^x)^y = a^{x \, \times \, y}}$

### Example Question 2

Write the following expression in the form $\boldsymbol{xa^y}$ where $\boldsymbol{a \neq 0}$:
$\boldsymbol {(\frac{3a^3 \, \times \, 4a^{-5}}{2a^4})^2}$

Start by using the power rule on the numerator:

$(\frac{3a^3 \, \times \, 4a^{-5}}{2a^4})^2 = (\frac{12a^{-2}}{2a^4})^2$

Then, use the division rule:

$(\frac{12a^{-2}}{2a^4})^2$ $= (6a^{-6})^2$

Finally, use the power rule:

$(6a^{-6})^2 = 36a^{-12}$

These rules also apply to algebraic indices:

### Example Question 3

Find the value of $\boldsymbol{b}$, if $\boldsymbol{\frac{3^{4x\,-\,2}}{3^{x\,-\,1}} \equiv 3^{ax+b}}$

$\frac{3^{4x\,-\,2}}{3^{x\,-\,1}} = 3^{(4x\,-\,2)\,-\,(x\,-\,1)} = 3^{3x\,-\,1}$

Therefore, $b$ = -1.

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