GCSE Maths · Number

Indices GCSE Questions, Worked Examples and Answers

An index is the small raised number in a power. In $5^3$ , there are three factors of the base $5$ : $5\times5\times5$ . This guide builds every index law from that meaning, then applies the laws to numerical and algebraic GCSE questions.

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Meaning before rules

What you need to know about indices

Index notation is a short way to record repeated multiplication. Once that meaning is secure, the laws describe what happens to the number of repeated factors.

The base is 5. It is the factor being repeated.

The index is 3. It tells you there are three factors of 5. The whole expression is a power, read as “five to the power of three” or “five cubed”.

$5^3=5\times5\times5=125$

Multiply

Joining factors makes indices add

$2^3\times2^4=(2\times2\times2)(2\times2\times2\times2)=2^7$

There are $3+4$ factors of 2, so $a^m\times a^n=a^{m+n}$ . Use this only when the operation is multiplication and the bases are the same.

Do not use it for: $2^3\times3^4$ or $x^3+x^4$ .

Divide

Cancelling factors makes indices subtract

$\frac{3^5}{3^2}=\frac{3\times3\times3\times3\times3}{3\times3}=3^3$

Two matching factors cancel, leaving $5-2$ factors. Therefore $a^m\div a^n=a^{m-n}$ for $a\ne0$ .

Do not subtract indices when the bases are different.

Power of a power

Equal groups make indices multiply

$(x^2)^3=(x^2)(x^2)(x^2)=x^6$

Three groups each contain two factors of $x$ , giving $2\times3$ factors. So $(a^m)^n=a^{mn}$ .

Brackets matter: $(2x^3)^2=4x^6$ , because the outside power also squares the 2.

Zero index

Why a non-zero base to power zero is 1

$\frac{7^4}{7^4}=1\qquad\text{but also}\qquad\frac{7^4}{7^4}=7^{4-4}=7^0$

Both expressions describe the same quotient, so $7^0=1$ . In general, $a^0=1$ for $a\ne0$ . This argument does not define $0^0$ .

$2^3=8$

÷ 2

$2^2=4$

÷ 2

$2^1=2$

÷ 2

$2^0=1$

÷ 2

$2^{-1}=\frac12$

Extend the same pattern

Negative and fractional indices

These are not separate tricks. Negative indices continue the divide-by-the-base pattern, while fractional indices undo whole-number powers.

A negative index means a reciprocal

$4^{-2}=\frac{1}{4^2}=\frac1{16}$

Moving from index $0$ to $-1$ divides by the base once; moving to $-2$ divides again. So $a^{-n}=\frac1{a^n}$ .

It is not negative: $4^{-2}\ne-16$ .

Higher only

A fractional index describes a root

$9^{\frac12}=3\quad\text{because}\quad(9^{\frac12})^2=9^1=9$

The denominator gives the root and the numerator gives the power:

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

For easy arithmetic, take the root first: $64^{\frac23}=(\sqrt[3]{64})^2=4^2=16$ .

Before applying an index law, ask

What is the operation? Multiplication, division and a power outside brackets use different laws.
Are the bases the same? The product and quotient laws need matching bases.
Is there a coefficient? Calculate coefficients separately and apply an outside power to every factor inside brackets.
What form is required? A negative power may need rewriting with positive indices; a numerical power may need evaluating.

Fully worked

Indices GCSE worked examples

The examples move from one law at a time to coefficients, fractional indices and multi-step problems.

Example 1

Multiply powers of the same base

1 mark

Question

Simplify $6^2\times6^5$ .

The base is 6 in both powers and the operation is multiplication, so join the factors by adding the indices.

$6^2\times6^5=6^{2+5}$

$\boxed{6^7}$

Exam tip: leave the result in index form when the command is “simplify”. Do not calculate $6^7$ unless asked.

Example 2

Divide algebraic powers

1 mark

Question

Simplify $\dfrac{p^9}{p^4}$ .

Matching factors cancel in a quotient, so subtract the denominator’s index from the numerator’s index.

$\frac{p^9}{p^4}=p^{9-4}$

$\boxed{p^5}$

Check: $p^5\times p^4=p^9$ , so the simplified quotient reverses correctly.

Example 3

Apply an outside power to every factor

2 marks

Question

Simplify $(3a^2)^3$ .

The outside power applies to the coefficient 3 and to $a^2$ .

$(3a^2)^3=3^3(a^2)^3$

$=27a^{2\times3}$

$\boxed{27a^6}$

Common loss of marks: writing $3a^6$ forgets that the coefficient is inside the brackets.

Example 4

Use zero and negative indices

3 marks

Question

Work out $4^0+2^{-3}$ . Give your answer as a fraction.

Evaluate each power before adding. The zero index produces 1; the negative index produces a reciprocal.

$4^0=1$

$2^{-3}=\frac1{2^3}=\frac18$

$1+\frac18=\frac98$

$\boxed{\frac98}$

Sign check: $2^{-3}$ is a small positive number, not $-8$ .

Example 5 · Higher only

Evaluate a fractional index

2 marks

Question

Work out $64^{\frac23}$ without a calculator.

The denominator 3 means cube root; the numerator 2 means square. Taking the root first keeps the numbers small.

$64^{\frac23}=\left(\sqrt[3]{64}\right)^2$

$=4^2$

$\boxed{16}$

Memory cue: denominator = root. It sits underneath the fraction just as a root sign sits over the number.

Example 6

Simplify coefficients and two bases

3 marks

Question

Simplify $\dfrac{18x^7y^3}{6x^2y}$ .

Treat the coefficient, the powers of $x$ and the powers of $y$ as three separate quotients.

$\frac{18}{6}=3$

$\frac{x^7}{x^2}=x^{7-2}=x^5$

$\frac{y^3}{y}=y^{3-1}=y^2$

$\boxed{3x^5y^2}$

Notice: an unwritten index is 1, so $y=y^1$ .

Example 7

Combine three index laws

4 marks

Question

Simplify $\dfrac{(2a^3)^2a^4}{4a^5}$ .

Start with the brackets, then multiply in the numerator, then divide. Keeping one transformation per line protects the coefficient.

$(2a^3)^2=4a^6$

$4a^6\times a^4=4a^{10}$

$\frac{4a^{10}}{4a^5}=a^{10-5}$

$\boxed{a^5}$

Order: deal with the outside power before using the multiplication and division laws.

Example 8 · Higher only

Combine fractional indices with different bases

4 marks

Question

Work out $\dfrac{16^{\frac32}}{8^{\frac23}}$ without a calculator.

The quotient law cannot be used while the bases are 16 and 8. Rewrite both as powers of 2, then apply the power-of-a-power law.

$16=2^4\qquad8=2^3$

$\frac{16^{\frac32}}{8^{\frac23}}=\frac{(2^4)^{\frac32}}{(2^3)^{\frac23}}$

$=\frac{2^6}{2^2}$

$=2^{6-2}=2^4$

$\boxed{16}$

Check another way: $16^{\frac32}=64$ and $8^{\frac23}=4$ , so the quotient is $64\div4=16$ .

15 original questions

Indices GCSE exam-style questions

Try each question before opening its worked answer. Higher-only content is labelled on the individual card.

Before you startAllow about 35 minutes · show each index-law step · total 29 marks

Write repeated multiplication in index form

1 mark

Write $7\times7\times7\times7\times7$ in index form.

Show worked answer

The repeated factor is $7$ , so $7$ is the base. It appears five times, so the index is $5$ .

$7\times7\times7\times7\times7=\boxed{7^5}$

Evaluate a power

1 mark

Work out $3^4$ .

Show worked answer

The index $4$ means that four factors of $3$ are multiplied.

$3^4=3\times3\times3\times3$

$=9\times9$

$=\boxed{81}$

Multiply powers of the same base

1 mark

Simplify $b^4\times b^7$ .

Show worked answer

Both powers have base $b$ . Multiplying joins four factors of $b$ to seven more, so add the indices.

$b^4\times b^7=b^{4+7}$

$=\boxed{b^{11}}$

Divide powers of the same base

1 mark

Simplify $\dfrac{c^{10}}{c^3}$ .

Show worked answer

Three factors of $c$ cancel from the numerator and denominator, leaving seven. This is why the indices are subtracted.

$\frac{c^{10}}{c^3}=c^{10-3}$

$=\boxed{c^7}$

Raise a power to a power

1 mark

Simplify $(m^3)^4$ .

Show worked answer

There are four groups of $m^3$ , so there are $4\times3$ factors of $m$ altogether.

$(m^3)^4=m^{3\times4}$

$=\boxed{m^{12}}$

Keep the coefficient

2 marks

Simplify $5x^2\times3x^6$ .

Show worked answer

Multiply the ordinary coefficients, then apply the product law only to the powers with base $x$ .

$5\times3=15$

$x^2\times x^6=x^{2+6}=x^8$

$\boxed{15x^8}$

Simplify an algebraic fraction

3 marks

Simplify $\dfrac{24a^8b^5}{6a^3b^2}$ .

Show worked answer

Deal with the coefficient and each base separately.

$\frac{24}{6}=4$

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

$\frac{b^5}{b^2}=b^{5-2}=b^3$

$\boxed{4a^5b^3}$

Use the zero index

1 mark

Work out $9^0$ .

Show worked answer

Any non-zero base to the power zero is $1$ . For example, $9^2\div9^2=9^{2-2}=9^0$ , while the fraction is also $1$ .

$\boxed{9^0=1}$

Evaluate a negative index

2 marks

Work out $5^{-3}$ . Give your answer as a fraction.

Show worked answer

A negative index means take the reciprocal of the corresponding positive power.

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

$5^3=125$

$\boxed{5^{-3}=\frac{1}{125}}$

The value is positive: the minus sign is part of the index, not a sign in front of the number.

Higher only

Connect fractional indices and roots

2 marks

Work out $81^{\frac12}+27^{\frac13}$ .

Show worked answer

The denominator of each fractional index gives the root.

$81^{\frac12}=\sqrt{81}=9$

$27^{\frac13}=\sqrt[3]{27}=3$

$9+3=\boxed{12}$

Higher only

Use a fractional index with a numerator

2 marks

Work out $32^{\frac35}$ without a calculator.

Show worked answer

Use the denominator $5$ as the root first, then use the numerator $3$ as the power.

$32^{\frac35}=\left(\sqrt[5]{32}\right)^3$

$=2^3$

$=\boxed{8}$

Higher only

Combine negative and fractional indices

3 marks

Work out $16^{-\frac34}$ without a calculator.

Show worked answer

First use the negative sign to write a reciprocal.

$16^{-\frac34}=\frac{1}{16^{\frac34}}$

Now take the fourth root before cubing.

$16^{\frac34}=\left(\sqrt[4]{16}\right)^3$

$=2^3=8$

$\boxed{16^{-\frac34}=\frac18}$

Combine several index laws

3 marks

Simplify $\dfrac{(3p^2)^2p^5}{9p^3}$ .

Show worked answer

Apply the outside power to both the coefficient and the power of $p$ .

$(3p^2)^2=3^2p^{2\times2}=9p^4$

Multiply the powers of $p$ in the numerator.

$9p^4\times p^5=9p^9$

Now divide the coefficient and subtract the indices.

$\frac{9p^9}{9p^3}=p^{9-3}$

$=\boxed{p^6}$

Explain why addition is different

2 marks

A student writes $x^3+x^3=x^6$ .

Explain the mistake and give the correct simplified expression.

Show worked answer

Adding indices is a rule for multiplying powers of the same base. Here the operation is addition, so the two identical terms are collected instead.

$x^3+x^3=1x^3+1x^3$

$=\boxed{2x^3}$

By contrast, $x^3\times x^3=x^{3+3}=x^6$ .

Higher only

Rewrite bases to solve an index equation

4 marks

Solve $9^{x+1}=27^{x-1}$ .

Show worked answer

Rewrite both bases as powers of $3$ .

$9=3^2\qquad\text{and}\qquad27=3^3$

$(3^2)^{x+1}=(3^3)^{x-1}$

Use the power-of-a-power law.

$3^{2(x+1)}=3^{3(x-1)}$

Equal powers with the same positive base have equal indices.

$2(x+1)=3(x-1)$

$2x+2=3x-3$

$\boxed{x=5}$

Check: both original sides become $3^{12}$ .

Protect your marks

Common indices mistakes

Most index errors come from applying a correct rule to the wrong operation or forgetting that a coefficient is a separate factor.

Multiplying the indices

$x^3\times x^4=x^7$ , not $x^{12}$ . Add because you are joining 3 and 4 factors. Multiply indices only for a power of a power.

Using index laws with addition

$x^4+x^4=2x^4$ , not $x^8$ . The add-the-indices law needs multiplication between the powers.

Treating a negative index as a negative value

$3^{-2}=\frac19$ , not $-9$ . The minus sign tells you to take the reciprocal.

Reversing a fractional index

In $a^{\frac mn}$ , the denominator $n$ is the root and the numerator $m$ is the power. Take the root first when it gives an integer.

Losing coefficients

$(4x^2)^3=64x^6$ , not $4x^6$ . An outside power applies to every factor inside the brackets.

Ignoring different bases

You cannot turn $2^3\times5^4$ into one power by adding indices. First check whether the bases match or can sensibly be rewritten with a common base.

30-second recap

Operation, base, coefficient, final form

Read the operation first. Check the bases. Handle coefficients separately. Then check whether your answer should have positive indices or a numerical value.

Multiplying same bases: add indices.
Dividing same bases: subtract indices.
Power of a power: multiply indices.
Zero gives 1; negative gives a reciprocal.
For a fractional index, denominator means root.

Quick answers

Indices FAQ

What are indices in GCSE Maths?

Indices are the small raised numbers used to write repeated multiplication compactly. In 5³, 5 is the base and 3 is the index, so the expression means three equal factors of 5.

What are the main index laws?

For the same base, add indices when multiplying and subtract indices when dividing. Multiply the indices when raising a power to another power. Each law follows from counting repeated factors, and none of them is a rule for adding terms.

Does a negative index make the value negative?

No. A negative index means take the reciprocal: a⁻ⁿ = 1/aⁿ for a non-zero base. For example, 3⁻² = 1/9, which is positive.

How are fractional indices connected to roots?

The denominator gives the root and the numerator gives the power. For example, 64 to the power of 2/3 means square the cube root of 64. Fractional indices are Higher-tier content.

Can I use index laws when powers are added?

No. The product law needs multiplication. For example, x³ + x³ = 2x³, whereas x³ × x³ = x⁶.

Build the connection

What to revise next

Indices connect directly to roots, exact values and calculations with powers of ten.

Higher extension

Fractional indices

Practise roots, general fractional powers and negative fractional indices in more depth.

Dedicated guide coming soon

Exact values

Surds

Use roots that do not simplify to integers and keep irrational values in exact form.

Dedicated guide coming soon

Apply index laws

Standard form

Calculate with very large and very small numbers using powers of 10.

Revise standard form

Content standards

Curriculum and rights review

Curriculum references checked 3 September 2026. This guide supports index notation and index laws used across Edexcel, AQA and OCR GCSE Maths. Fractional indices and index equations are labelled Higher only where they appear. All questions, explanations and worked solutions are original Pass an Exam material; no past-paper wording has been reproduced.

Official specification references