Indices Rules Worksheets | Questions and Revision | MME

# Indices Rules Worksheets, Questions and Revision

Level 6-7

## Indices Rules

Indices Rules builds on the 7 rules from Powers and Roots. We will cover 3 more complicated rules here. Make sure you are confident with the following topics before moving onto laws and indices

Level 6-7

## Indices Rule 8: Fractional Powers

The fractional indices laws apply when the power is a fraction.

$\textcolor{red}{a}^{\large{\frac{\textcolor{blue}{b}}{\textcolor{limegreen}{c}}}} = \sqrt[\textcolor{limegreen}{c}]{\textcolor{red}{a}^\textcolor{blue}{b}}$

This is commonly use to show square and cube roots

$\textcolor{red}{x}^{\large{\frac{\textcolor{limegreen}{1}}{\textcolor{blue}{2}}}}= \sqrt[\textcolor{blue}{2}]{\textcolor{red}{x}^\textcolor{limegreen}{1}} =\sqrt[\textcolor{blue}{2}]{\textcolor{red}{x}}$

$\textcolor{red}{x}^{\large{\frac{\textcolor{limegreen}{1}}{\textcolor{blue}{3}}}}= \sqrt[\textcolor{blue}{3}]{\textcolor{red}{x}^\textcolor{limegreen}{1}} =\sqrt[\textcolor{blue}{3}]{\textcolor{red}{x}}$

Note: it doesn’t matter which order you carry out the square root and multiplication operations. In other words, the rule can also be written as

$\textcolor{red}{a}^{\large{\frac{\textcolor{blue}{b}}{\textcolor{limegreen}{c}}}} = (\sqrt[\textcolor{limegreen}{c}]{\textcolor{red}{a}})^\textcolor{blue}{b}$

You should try to carry out the operations in the order that makes the calculation as simple as possible.

Level 6-7

## Indices Rule 9: Multi-step Fractional Powers

You may also be asked to simplify expressions where the numerator is not $\bf{1}$

$\textcolor{red}{64}^{\large{\frac{\textcolor{limegreen}{2}}{\textcolor{blue}{3}}}}= \sqrt[\textcolor{blue}{3}]{\textcolor{red}{64}^\textcolor{limegreen}{2}}$

$\sqrt[\textcolor{blue}{3}]{\textcolor{red}{64}} = \textcolor{red}{4}$

$\textcolor{red}{4}^\textcolor{limegreen}{2} = \textcolor{red}{16}$

Level 6-7

## Indices Rule 10: Negative Powers

Negative powers flip the fraction and put $1$ over the number

In general, the result of a negative power is “$\bf{1}$ over that number to the positive power”, i.e.

$\textcolor{red}{a}^{-\textcolor{limegreen}{b}} = \dfrac{1}{\textcolor{red}{a}^\textcolor{limegreen}{b}}$

for any value of $a$ or $b$. When the power is $\textcolor{blue}{-1}$, this takes the form,

$\textcolor{red}{a}^{\textcolor{blue}{-1}}=\dfrac{1}{\textcolor{red}{a}}$ or $\textcolor{red}{10}^{\textcolor{blue}{-1}} = \dfrac{1}{\textcolor{red}{10}}$

When the number is a fraction, the negative power flips the fraction

$\bigg(\dfrac{\textcolor{blue}{a}}{\textcolor{limegreen}{b}}\bigg)^{-\textcolor{red}{x}} = \bigg(\dfrac{\textcolor{limegreen}{b}}{\textcolor{blue}{a}}\bigg)^\textcolor{red}{x}$

Level 6-7

## Example 1: Negative Powers

Simplify the following, $4^{-3}$.

[2 marks]

We now know that $4^{-3}$ is equal to $\dfrac{1}{4^3}$. We also know that

$4^3=4\times 4\times 4=16\times 4=64$.

So, we get that

$4^{-3}=\frac{1}{64}$.

Level 6-7

## Example 2: Fractional Powers and Roots

Simplify the following, $9^{\frac{3}{2}}$.

[2 marks]

So, we know that $9^{\frac{3}{2}}$ is equal to $\sqrt[2]{9^3}$ or $(\sqrt[2]{9})^3$

So, to work out $(\sqrt[2]{9})^3$, we first have to square root $9$, which is easy enough – the square root of $9$ is $3$. So, $(\sqrt[2]{9})^3$ becomes $3^3$, which is

$3^3=3\times 3\times 3 = 27$

Level 6-7

## Example 3: Multiplication and Powers

Write $2^{15}\times 8^{-4}$ as a power of $2$, and hence evaluate the expression. (Non calculator)

[3 marks]

The first part of the expression is a power of $2$, whilst the second part is a power of $8$.

we know that

$8 = 2^3$

This means we can rewrite the following,

$8^{-4}=\left(2^3\right)^{-4}$

Next, using Rule 3, we can simplify,

$\left(2^3\right)^{-4}=2^{3\times(-4)}=2^{-12}$

So the whole expression can be written as

$2^{15}\times2^{-12},$

Finally using Rule 1 we simplify the expression further.

$2^{15}\times2^{-12}=2^{15+(-12)}=2^3$

Thus, we have written the expression as a power of $2$. Evaluating this final answer gives

$2^3 = 8$

Level 6-7

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### Example Questions

So, we can’t use any laws straight away since the terms don’t have the same base. However, if we recognise that $9=3^2$, then we can write the first term as

$\left(3^2\right)^5$

Using the power law, we get

$\left(3^2\right)^5=3^{2\times5}=3^{10}$

Therefore, the whole expression becomes

$3^{10}\times3^{-5}$

Applying the multiplication law, this simplifies to

$3^{10+(-5)}=3^5$

Thus, we have written the expression as a power of $3$.

Firstly, as $3^2=9$, the inverse operation gives, $\sqrt{9}=3$

So, that leaves $6^{-2}$, this becomes the following fraction,

$6^{-2}=\dfrac{1}{6^2}$

We know that $6^2=6\times 6=36$, so

$6^{-2}=\dfrac{1}{36}$

Multiplying our two answers together, we get

$\sqrt{9}\times 6^{-2}=3\times\dfrac{1}{36}=\dfrac{3}{36}=\dfrac{1}{12}$

This expression can be rewritten as,

$\sqrt4 \times (\sqrt4)^3$

Given we know that $\sqrt4=2$ , this becomes,

$2\times2^3$

Hence,

$2\times2^3=2\times8=16$

Notice that in this example we chose to perform the $\sqrt{4}$ operation before cubing the answer. We could alternatively write the expression as $\sqrt{4^3}$, but in this case the first option is easier.

As it is a negative power we can rewrite this as,

$8^{-\frac{5}{3}}=\frac{1}{8^{\frac{5}{3}}}$

Now, we can work out the denominator, which we will write as,

$8^{\frac{5}{3}}=\sqrt[3]{8^5}=(\sqrt[3]{8})^5$

We know that $\sqrt[3]{8}=2$. So this simplifies to,

$(\sqrt[3]{8})^5=2^5$

Counting up in powers of $2$: $4$, $8$, $16$, $32$ – we see that $32$ is the $5$th power of $2$, so

$\sqrt[3]{8}^5=32$

$8^{-\frac{5}{3}}=\dfrac{1}{32}$

### Worksheets and Exam Questions

#### (NEW) Rules of Indices Exam Style Questions - MME

Level 6-7 New Official MME

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