Index Laws

Syllabus

use index laws and surds

Index Notation

Index notation is used to represent expressions that deal with numbers that are repeatedly multiplied together.

If $n$ is a positive integer, then $a^n=\underbrace{a\times a\times a\times \dots\times a}_{n\text{ factors}}$.

$a$ is known as the base and $n$ is known as the power, index or exponent.

Index Laws

You should be familiar with the following index laws from previous years.

Important

• $a^m\times a^n=a^{m+n}$

• $\dfrac{a^m}{a^n}=a^{m-n}$

• $\left(a^m\right)^n=a^{mn}$

• $(ab)^m=a^mb^m$

• $\left(\dfrac{a}{b}\right)^m=\dfrac{a^m}{b^m}$

• $a^0=1, a\neq0$

• $a^{-m}=\dfrac{1}{a^m}$

• $a^{\frac{m}{n}}=\left(\sqrt[n]{a}\right)^m=\sqrt[n]{a^m}$

You should recall from the last index law listed above, surds can be expressed as indices and vice versa.

Example 1

Simplify the following:

1. $6^a\times36$

2. $\dfrac{\left(2a^2bc\right)^3}{bc^2}$

Solution

Solution 1

$$\begin{aligned} 6^a\times36&=6^a\times6^2&&\text{(express both numbers in terms of the same base)}\\ &=6^{a+2}&&\text{(add indices when multiplying numbers with equal base)} \end{aligned}$$

Solution 2

$$\begin{aligned} \dfrac{\left(2a^2bc\right)^3}{bc^2}&=\dfrac{2^3\left(a^2\right)^3b^3c^3}{bc^2}&&\text{(remove brackets by raising each factor inside by the index)}\\ &=\dfrac{8a^6b^3c^3}{bc^2}&&\text{(multiply the indices when raising a power to a power)}\\ &=8a^6b^2c&&\text{(divide common factors)} \end{aligned}$$

Example 2

Simplify the following leaving your answer without brackets and with positive indices:

1. $\left(\dfrac{5a}{2b}\right)^3$

2. $\left(3\dfrac{1}{3}\right)^{-2}$

3. $\left(\dfrac{2a^{-2}}{3b}\right)^{-3}$

Solution

Solution 1

$$\begin{aligned} \left(\dfrac{5a}{2b}\right)^3&=\dfrac{5^3a^3}{2^3b^3}&&\text{(remove brackets by raising each factor inside by the index)}\\ &=\dfrac{125a^3}{8b^3}&&\text{(simplify)} \end{aligned}$$

Solution 2

$$\begin{aligned} \left(3\dfrac{1}{3}\right)^{-2}&=\left(\dfrac{10}{3}\right)^{-2}&&\text{(express the mixed numeral as an improper fraction)}\\ &=\dfrac{10^{-2}}{3^{-2}}&&\text{(remove brackets by raising each factor inside by the index)}\\ &=\dfrac{3^2}{10^2}&&\text{(take the reciprocal of any factor with a negative index)}\\ &=\dfrac{9}{100}&&\text{(simplify)} \end{aligned}$$

Solution 3

$$\begin{aligned} \left(\dfrac{2a^{-2}}{3b}\right)^{-3}&=\left(\dfrac{2}{3a^2b}\right)^{-3}&&\text{(take the reciprocal of any factor with a negative index)}\\ &&&\text{Note: this could be done later but it is good practice to simplify brackets first}\\ &=\dfrac{2^{-3}}{3^{-3}\left(a^2\right)^{-3}b^{-3}}&&\text{(remove brackets by raising each factor inside by the index)}\\ &=\dfrac{3^3\left(a^2\right)^3b^3}{2^3}&&\text{(take the reciprocal of any factor with a negative index)}\\ &=\dfrac{27a^6b^3}{8}&&\text{(multiply the indices when raising a power to a power and simplify)}\\ \end{aligned}$$

Example 3

Express the following without a fraction using a prime number base:

1. $\dfrac{27^{a+1}}{9^{a-1}}$

2. $\dfrac{1}{\sqrt[6]{243}}$

Solution

Solution 1

$$\begin{aligned} \dfrac{27^{a+1}}{9^{a-1}}&=\dfrac{\left(3^3\right)^{a+1}}{\left(3^2\right)^{a-1}}&&\text{(express both numbers in terms of the same base)}\\ &=\dfrac{3^{3(a+1)}}{3^{2(a-1)}}&&\text{(multiply the indices when raising a power to a power)}\\ &=3^{3(a+1)-2(a-1)}&&\text{(subtract indices when dividing numbers with equal base)}\\ &=3^{3a+3-2a+2}&&\text{(expand brackets)}\\ &=3^{a+5}&&\text{(collect like terms)} \end{aligned}$$

Solution 2

$$\begin{aligned} \dfrac{1}{\sqrt[6]{243}}&=\dfrac{1}{\sqrt[6]{3^5}}&&\text{(express 243 as a product of prime factors)}\\ &&&\text{\htmlClass{casio}{243=qx}}\\ &=\dfrac{1}{3^{\frac{5}{6}}}&&\text{(convert surd to fractional index)}\\ &=3^{-\frac{5}{6}}&&\text{(take reciprocal using negative index)} \end{aligned}$$