Indices Rules Worksheets | Questions and Revision | MME

Indices Rules Worksheets, Questions and Revision

Level 6-7
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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

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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 5th power of 2, so

 

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

 

Therefore, the answer is,

 

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

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Worksheets and Exam Questions

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(NEW) Rules of Indices Exam Style Questions - MME

Level 6-7 New Official MME

Drill Questions

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Indices Rules - Drill Questions

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Fractional And Negative Indices - Drill Questions

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Rules of Indices - Drill Questions

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