# Rules of Indices

## Rules of Indices Revision

**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**.

**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.

**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}

**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}

**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}.

**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

**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

## Rules of Indices Example Questions

**Question 1:** Write 9^5\times3^{-5} as a power of 3

**[3 marks]**

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.

**Question 2:** Work out \sqrt{9}\times 6^{-2}

Write your answer in its simplest form (Non-calculator)

**[3 marks]**

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}

**Question 3:** Work out 4^{\frac{1}{2}}\times4^{\frac{3}{2}}

**[3 marks]**

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.

**Question 4:** Work out 8^{-\frac{5}{3}} (HIGHER ONLY)

(Non-calculator)

**[3 marks]**

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 5th power of 2, so

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

Therefore, the answer is,

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