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National 5 Maths: Indices

Course content

  • Simplifying expressions using the laws of indices
  • Multiplication and division using positive and negative indices including fractions
  • Power of a product: \({(ab)}^m=a^mb^m\)
  • Power of a power: \((a^m)^n=a^{mn}\)
  • Understanding the link between fractional indices and surds: \(a^{m/n}=\sqrt[\leftroot{-1}\uproot{3}\scriptstyle n]{a^m}\)

Key ideas

  • The word "index" means "power". For example: in 53, 5 is the "base" and 3 is the "index".
  • The plural of "index" is "indices".
  • Indices show repeated multiplication, eg. 53 = 5 \(\times\) 5 \(\times\) 5.
  • All of the other index laws are based on the simple facts above. The page below will explain why.

Textbook page references


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Laws of indices

There is no agreed numbering system for the index laws. We have decided to order them so that you can read this page from top to bottom. Each law should make sense because of what you have already read.

We will use the following numbering system and names for each law.

$$ \begin{eqnarray} &&\small\textsf{Law 1:}\normalsize&\:&a^{m}a^n=\ a^{m+n}&\:&\small\textsf{Multiplication}\normalsize\\[10pt] &&\small\textsf{Law 2:}\normalsize&\:&\left(a^m\right)^n=a^{mn}&\:&\small\textsf{Power of a power}\normalsize\\[6pt] &&\small\textsf{Law 3:}\normalsize&\:&\frac{a^m}{a^n} =\ a^{m-n}&\:&\small\textsf{Division}\normalsize\\[6pt] &&\small\textsf{Law 4:}\normalsize&\:&a^0=1&\:&\small\textsf{Power zero}\normalsize\\[6pt] &&\small\textsf{Law 5:}\normalsize&\:&a^{-m}=\frac{1}{a^m}&\:&\small\textsf{Negative power}\normalsize\\[6pt] &&\small\textsf{Law 6:}\normalsize&\:&a^{1/n}=\sqrt[\leftroot{-1}\uproot{2}\scriptstyle n]{a}&\:&\small\textsf{Unitary fraction}\normalsize\\[10pt] &&\small\textsf{Law 7:}\normalsize&\:&a^{m/n}=\sqrt[\leftroot{-1}\uproot{5}\scriptstyle n]{a^m}=(\sqrt[\leftroot{-1}\uproot{2}\scriptstyle n]{a})^m&\:&\small\textsf{General fraction}\normalsize\\[6pt] \end{eqnarray} $$

Law 1: Multiplication

Example:
$$ \begin{eqnarray} \ 7^3 \times 7^2 &=& \underbrace{(7\times7\times7)\times(7\times7)}_{\textsf{5 factors}}\\[6pt] &=& 7^5 \end{eqnarray} $$

Instead of multiplying this out in full, we can just add the indices: 3 + 2 = 5.

So we get the general law:

$$\large\boxed{a^{m}a^n=\ a^{m+n}}\normalsize$$

Law 2: Power of a power

Example:
$$ \begin{eqnarray} \ {(5^2)}^3 &=& 5^2\times5^2\times5^2\\[10pt] &=& \underbrace{(5\times5)\times(5\times5)\times(5\times5)}_{\textsf{6 factors}}\\[6pt] &=& 5^6 \end{eqnarray} $$

Instead of writing this out in full, we can just multiply the indices: 2 \(\times\) 3 = 6.

So we get the general law:

$$\large\boxed{\left(a^m\right)^n=a^{mn}}\normalsize$$

Law 3: Division

Example:
$$ \begin{eqnarray} 9^5\ \div\ 9^2 &=& \frac{9\times9\times9\times9\times9}{9\times9}\\[6pt] &=&\ \frac{9\times9\times9\times\cancel{9}\times\cancel{9}}{\cancel{9}\times\cancel{9}}\\[8pt] &=&\ 9\times9\times9\\[8pt] &=& 9^3 \end{eqnarray} $$

Instead of writing this out in full, we can just subtract the indices: 5 – 2 = 3.

So we get the general law:

$$\large\boxed{\frac{a^m}{a^n} =\ a^{m-n}}\normalsize$$

Law 4: Power zero

Let's think about this example: 58 ÷ 58

Law 3, above, says that 58 ÷ 58 = 58–8 = 50

But we also know that anything divided by itself equals 1. So 58 ÷ 58 = 1

This means that 50 must equal 1.

There was nothing special about the numbers 5 or 8 here. So we get the general law:

$$\large\boxed{a^0=1}\normalsize$$

Law 5: Negative power

Let's think about this example: 4–7

The negative number –7 means 7 below zero. In other words, –7 = 0 – 7.

Now we can use Law 3 backwards:
$$ \begin{eqnarray} 4^{-7} &=&\ 4^{0-7}\\[6pt] &=&\ \frac{4^0}{4^7}\\[6pt] &=&\ \frac{1}{4^7} \end{eqnarray} $$

There was nothing special about the numbers 4 or 7 here. So we get the general law:

$$\large\boxed{a^{-m}=\frac{1}{a^m}}\normalsize$$

Law 6: Unitary fraction

"Unitary" means that the numerator is 1.

As an example, let's try to work out what \(9^{\frac{1}{2}}\) means.

If we multiply it by itself, we get \(9^{\frac{1}{2}}\times9^{\frac{1}{2}}\ =\ 9^{\frac{1}{2}+\frac{1}{2}}\ =\ 9^1\ =\ 9\).

But we also know that \(\sqrt9\ \times\sqrt{9}\ =\ 9\). This means that \(9^{\frac{1}{2}}\) and \(\sqrt{9}\) are the same thing.

Similarly, power \(\frac{1}{3}\) means cube root, power \(\frac{1}{4}\) means 4th root, and so on.

So we get the general law:

$$\large\boxed{a^{1/n}=\sqrt[\leftroot{-1}\uproot{2}\scriptstyle n]{a}}\normalsize$$

Law 7: General fraction

This final law follows on from Law 6 above.

First, note that \(\frac{m}{n}\ =\ m\times\frac{1}{n}\ =\ \frac{1}{n}\times m\)

If we split \(\frac{m}{n}\) the first way, we get:

$$ \begin{eqnarray} a^{m/n}\ &=&\ a^{m\times{\frac{1}{n}}}\\[6pt] &=&\ (a^m)^{\frac{1}{n}}\ \ \small\textsf{(using Law 2)}\normalsize\\[6pt] &=&\ \sqrt[\leftroot{-1}\uproot{6}\scriptstyle n]{a^m} \end{eqnarray} $$

If we split \(\frac{m}{n}\) the second way, we get:

$$ \begin{eqnarray} a^{m/n}\ &=&\ a^{{\frac{1}{n}}\times m}\\[6pt] &=&\ (a^{\frac{1}{n}})^m\ \ \small\textsf{(using Law 2)}\normalsize\\[6pt] &=&\ (\sqrt[\leftroot{-1}\uproot{2}\scriptstyle n]{a})^m \end{eqnarray} $$

So we get the general law:

$$\large\boxed{a^{m/n}=\sqrt[\leftroot{-1}\uproot{5}\scriptstyle n]{a^m}=(\sqrt[\leftroot{-1}\uproot{1}\scriptstyle n]{a})^m}\normalsize$$

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Example 1 (calculator)

SQA National 5 Maths 2014 P2 Q8

Simplify \( \large\frac{n^5 \times 10n}{2n^2}\small. \)

Example 2 (non-calculator)

SQA National 5 Maths 2015 P1 Q14

Evaluate \( 8^{\frac53}\small. \)

Example 3 (calculator)

SQA National 5 Maths 2016 P2 Q10

Simplify \( (n^2)^3\times n^{-10}\small.\)
Give your answer with a positive power.

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Example 4 (calculator)

SQA National 5 Maths 2017 P2 Q12

Express \( \large\frac{1}{\sqrt[\leftroot{-1}\uproot{2} 3]{x}}\) in the form \(x^{n}\small.\)

Example 5 (non-calculator)

SQA National 5 Maths 2018 P1 Q15

Remove the brackets and simplify \( \left(\frac{2}{3}p^4\right)^2\small. \)

Example 6 (non-calculator)

Simplify, expessing your answer with a positive power: \( \left(\frac{2}{3}p^{-4}\right)^2\small. \)

Example 7 (calculator)

SQA National 5 Maths 2019 P2 Q16

Simplify \( \large\frac{a^4 \times 3a}{\sqrt{a}}\small. \)

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Example 8 (non-calculator)

SQA National 5 Maths 2021 P1 Q15

Evaluate \( 16^{\frac32}\small. \)

Example 9 (non-calculator)

SQA National 5 Maths 2022 P1 Q11

Simplify \( (m^{-2})^4\times m^{-5}\small.\)
Give your answer with a positive power.

Example 10 (non-calculator)

SQA National 5 Maths 2023 P1 Q12

Simplify \( \large\frac{5c^{-2}}{c^3 \times c^4}\small. \)
Give your answer with a positive power.

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Maths.scot worksheet

• Indices worksheet
• Answer sheet
• See all National 5 Maths worksheets

Past paper questions

• All past paper questions by topic
• 2013 Spec. P1 Q7 (with expansion)
• 2014 Paper 2 Q8
• 2015 Paper 1 Q14 (evaluation)
• 2016 Paper 2 Q10
• 2017 Paper 2 Q12
• 2018 Paper 1 Q15
• 2019 Paper 2 Q16
• 2021 Paper 1 Q15 (evaluation)
• 2022 Paper 1 Q11
• 2023 Paper 1 Q12

Other great resources

Videos - Maths180.com
Videos - Larbert High School
1. Multiplication
2. Power of a power
3. Division
4. Non-unitary fractions
5. Negative powers
Videos - Mr Graham Maths
1. Multiplication
2. Power of a power
3. Division
4. Negative powers
5. Fractions
Videos - Mr Murray Maths Help
1. Multiplication
2. Power of a power
3. Division
4. Power 0
5. Negative powers
6. Unitary fractions
7. Non-unitary fractions
PowerPoint - MathsRevision.com
Worked examples - Maths Mutt
Notes - BBC Bitesize
Notes - Maths4Scotland
1. Summary of index laws
2. Detailed notes
Notes - National5.com
Lesson notes - Maths 777
1. Multiplying and dividing
2. Powers of powers
3. Fractional indices
4. Brackets with indices
Practice questions - Maths Hunter
Worksheet - Airdrie Academy
Essential Skills worksheets
• Multiplying brackets (Answers)
• Fractional indices (Answers)
• Simplifying indices (Answers)
Worksheets - Calderglen HS
Combined with the surds topic
Exercises - Larkhall Academy
Pages 9-17 Ex 1-9 (no answers)
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