N6, N7NumberFoundation & HigherGrades 2–7

Indices & Roots

Q: Does 25 equal 5 or 5?

The symbol 25 means the positive root, 5. But the equation x^2 = 25 has two solutions, x = 5.

Q: Can I add indices when adding powers, like 2^3 + 2^4?

No. The laws are for multiplying and dividing only. 2^3 + 2^4 = 8 + 16 = 24, which is not 2^7.

An index (or power) tells you how many times to multiply a number by itself: $2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32$ . Roots undo powers. Three short laws let you multiply, divide and raise powers without working them out, and they extend naturally to zero, negative and fractional indices.

45 min Video by Maths Genie AQA GCSE Maths

The big picture

The laws of indices are not rules to memorise blindly — they come from counting. $3^4 \times 3^2$ is four 3s multiplied by two more 3s, so six 3s: $3^6$ . Division cancels factors, so $3^6 \div 3^2$ leaves four: $3^4$ .

Keep the pattern going and the strange-looking cases make sense. $3^3, 3^2, 3^1$ go $27, 9, 3$ — dividing by 3 each time — so $3^0 = 1$ and $3^{-1} = \frac{1}{3}$ . A negative index means a reciprocal, never a negative number.

Indices (Laws of Indices)Curated by Kepler, not produced by us. Watch it, then read the notes below.

Open on YouTube

What you'll learn

Calculate squares, cubes, square roots and cube roots
Apply the laws of indices to multiply, divide and raise powers
Evaluate zero and negative indices
Evaluate fractional indices (Higher)

Powers and roots

Learn the squares to $15^2 = 225$ and the cubes $1, 8, 27, 64, 125$ and $1000$ — they appear constantly on the non-calculator paper.

The square root $\sqrt{x}$ is the positive number that squares to $x$ ; every positive number also has a negative square root. The cube root $\sqrt[3]{x}$ can be negative: $\sqrt[3]{-8} = -2$ .

Example 1Work out

\sqrt{144} + 2^4 - \sqrt[3]{27}

\sqrt{144} = 12

2^4 = 16

\sqrt[3]{27} = 3

12 + 16 - 3

Answer

25

Your turnSame method, new numbers — try it before moving on.

Work out $5^3 - \sqrt{169}$ .

Your answer

Show worked solution

Solution

1 $5^3 = 125$ and $\sqrt{169} = 13$ .
2 $125 - 13 = 112$ .

Answer: $112$

Laws of indices

When multiplying powers of the same base, $\textbf{add}$ the indices. When dividing, $\textbf{subtract}$ them. For a power of a power, $\textbf{multiply}$ them.

The laws only work when the base is the same: $2^3 \times 5^2$ cannot be combined into a single power.

Example 1Simplify, leaving as a power of 7: (a)

7^5 \times 7^3

(b)

7^{10} \div 7^4

(c)

(7^2)^5

1(a) Add:

5 + 3 = 8

2(b) Subtract:

10 - 4 = 6

3(c) Multiply:

2 \times 5 = 10

Answer

(a)

7^8

(b)

7^6

(c)

7^{10}

Your turnSame method, new numbers — try it before moving on.

Simplify $\dfrac{4^6 \times 4^3}{4^5}$ , giving your answer as a power of 4.

Your answer

Type it like 4^7.

Show worked solution

Solution

1Numerator: $4^{6 + 3} = 4^9$ .
2Divide: $4^{9 - 5} = 4^4$ .

Answer: $4^4$

Example 2Simplify

3x^4 \times 5x^2

1Multiply the numbers:

15

2Add the indices:

x^{4 + 2} = x^6

Answer

15x^6

Your turnSame method, new numbers — try it before moving on.

Simplify $(2y^3)^4$ .

Your answer

Type it like 8y^6.

Show hint

Both the 2 and the $y^3$ are raised to the power 4.

Show worked solution

Solution

1 $2^4 = 16$ .
2 $(y^3)^4 = y^{12}$ , so $16y^{12}$ .

Answer: $16y^{12}$

Tip — $(2y^3)^4$ is not $2y^{12}$ — the bracket means the 2 is raised to the power too.

Zero and negative indices

Any non-zero number to the power 0 is 1: $9^0 = 1$ .

A negative index means "one over": $a^{-n} = \frac{1}{a^n}$ . So $2^{-3} = \frac{1}{8}$ , and $\left(\frac{2}{3}\right)^{-2} = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$ .

Example 1Work out (a)

5^{-2}

(b)

\left(\frac{1}{4}\right)^{-1}

(c)

10^0 \times 10^{-3}

1(a)

\frac{1}{5^2} = \frac{1}{25}

2(b) The reciprocal of

\frac{1}{4}

is 4.

3(c)

1 \times \frac{1}{1000}

Answer

(a)

\frac{1}{25}

(b)

4

(c)

0.001

Your turnSame method, new numbers — try it before moving on.

Work out $4^{-3}$ as a fraction.

Your answer

Type a fraction like 1/8.

Show worked solution

Solution

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

Answer: $\frac{1}{64}$

Why $a^0 = 1$

Using the division law, $5^2 \div 5^2 = 5^0$ . But anything divided by itself is 1 — so $5^0$ has to equal 1. The index laws force these definitions.

Fractional indices (Higher)

The denominator of a fractional index is a root: $a^{\frac{1}{n}} = \sqrt[n]{a}$ . The numerator is a power: $a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^m$ .

Take the root first — it keeps the numbers small. A negative sign still means reciprocal.

Example 1Work out (a)

27^{\frac{2}{3}}

(b)

16^{-\frac{3}{4}}

1(a) Cube root:

\sqrt[3]{27} = 3

. Square:

9

2(b) Fourth root:

\sqrt[4]{16} = 2

. Cube:

8

. Reciprocal:

\frac{1}{8}

Answer

(a)

9

(b)

\frac{1}{8}

Your turnSame method, new numbers — try it before moving on.

Work out $32^{\frac{3}{5}}$ .

Your answer

Show worked solution

Solution

1Fifth root: $\sqrt[5]{32} = 2$ .
2Cube: $2^3 = 8$ .

Answer: $8$

Example 2Work out

\left(\frac{4}{9}\right)^{-\frac{3}{2}}

1Reciprocal:

\left(\frac{9}{4}\right)^{\frac{3}{2}}

2Square root:

\frac{3}{2}

. Cube:

\frac{27}{8}

Answer

\frac{27}{8}

Your turnSame method, new numbers — try it before moving on.

Work out $\left(\frac{8}{125}\right)^{-\frac{2}{3}}$ .

Your answer

Type a fraction like 9/4.

Show hint

Flip the fraction, cube root, then square.

Show worked solution

Solution

1Reciprocal: $\left(\frac{125}{8}\right)^{\frac{2}{3}}$ .
2Cube root: $\frac{5}{2}$ . Square: $\frac{25}{4}$ .

Answer: $\frac{25}{4}$

Think like an examiner

"Write as a single power of 5" means an answer like $5^9$ — evaluating it to a large number may not score.
Show the root and power steps for fractional indices: $\sqrt[3]{27} = 3$ , then $3^2 = 9$ .
Watch for questions mixing laws, e.g. $\frac{2^5 \times 2^{-2}}{2^4}$ — work out the index step by step.

Laws of indices

a^m \times a^n = a^{m+n}, \quad a^m \div a^n = a^{m-n}

Same base: add or subtract.

(a^m)^n = a^{mn}, \quad a^0 = 1

Power of a power; zero index.

a^{-n} = \dfrac{1}{a^n}, \quad a^{\frac{m}{n}} = (\sqrt[n]{a})^m

Negative and fractional indices.

Watch out for these

3^4 = 12

3^4 = 3 \times 3 \times 3 \times 3 = 81

2^{-3} = -8

A negative index is a reciprocal:

\frac{1}{8}

5^3 \times 5^2 = 25^5

Keep the base:

5^5

7^0 = 0

7^0 = 1

Stretch yourself

Find the value of $n$ if $\dfrac{9^n \times 3^5}{27} = 3^{12}$ .

Hint — Write every term as a power of 3: $9 = 3^2$ and $27 = 3^3$ .

Reveal solution

Solution

9^n = 3^{2n}

and

27 = 3^3

, so the left side is

3^{2n + 5 - 3} = 3^{2n + 2}

. Then

2n + 2 = 12

, so

n = 5

\blacksquare

Questions students ask

Does

\sqrt{25}

equal 5 or

\pm 5

The symbol

\sqrt{25}

means the positive root, 5. But the equation

x^2 = 25

has two solutions,

x = \pm 5

Can I add indices when adding powers, like

2^3 + 2^4

No. The laws are for multiplying and dividing only.

2^3 + 2^4 = 8 + 16 = 24

, which is not

2^7

What is

0^0

It is left undefined at GCSE, and it never appears in exam questions. Any non-zero number to the power 0 is 1.

Key takeaways

Know squares to 15 and cubes of 1–5 and 10.
Multiply: add indices. Divide: subtract. Power of a power: multiply.
$a^0 = 1$ and $a^{-n} = \frac{1}{a^n}$ .
$a^{\frac{m}{n}}$ : root with $n$ , power with $m$ .
Laws only combine powers with the same base.

Exam-style questions

Practise it the way it is examined

3 original questions · 9 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 33 marks

(a)Write $6^4 \times 6^7$ as a single power of 6.[1]
Show mark scheme (a)
- B1 $6^{11}$
Answer: $6^{11}$
(b)Write $\dfrac{2^9}{2^3 \times 2^2}$ as a single power of 2.[1]
Show mark scheme (b)
- B1 $2^4$
Answer: $2^4$
(c)Work out $8^0 + 8^{-1}$ .[1]
Show mark scheme (c)
- B1 $1\frac{1}{8}$ or $\frac{9}{8}$ or $1.125$
Answer: $\frac{9}{8}$

Question 2Grade 63 marks

Work out the value of $\left(\frac{81}{16}\right)^{-\frac{3}{4}}$ .[3]
Show mark scheme
- M1Deals with the negative: $\left(\frac{16}{81}\right)^{\frac{3}{4}}$ , or finds $\sqrt[4]{\frac{81}{16}} = \frac{3}{2}$
- M1Deals with the fourth root and the cube correctly
- A1 $\frac{8}{27}$
Answer: $\frac{8}{27}$

Question 3Grade 73 marks

$2^x \times 4^y = 2^{11}$ and $\dfrac{8^x}{2^y} = 2^{12}$ . Find $x$ and $y$ .[3]
Show mark scheme
- M1Writes in base 2: $x + 2y = 11$
- M1 $3x - y = 12$
- A1 $x = 5$ , $y = 3$
Answer: $x = 5$ , $y = 3$

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

0 / 0 first-time correct Streak 0Best at this level: 0

Generating a question…

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

Work out $3^4$ .
- $81$
- $12$
- $64$
- $27$
Show answer and explanation
Answer: $81$
$3 \times 3 \times 3 \times 3$ .
Work out $\sqrt[3]{125}$ .
- $5$
- $25$
- $41.6$
- $15$
Show answer and explanation
Answer: $5$
$5^3 = 125$ .
Simplify $x^5 \times x^4$ .
- $x^9$
- $x^{20}$
- $2x^9$
- $x^1$
Show answer and explanation
Answer: $x^9$
Add the indices.
Simplify $(a^3)^4$ .
- $a^{12}$
- $a^7$
- $a^{81}$
- $4a^3$
Show answer and explanation
Answer: $a^{12}$
Multiply the indices.
Work out $10^{-2}$ .
- $0.01$
- $-100$
- $-20$
- $0.1$
Show answer and explanation
Answer: $0.01$
$\frac{1}{100}$ .
Simplify $\frac{12y^8}{3y^2}$ .
- $4y^6$
- $4y^4$
- $9y^6$
- $36y^{10}$
Show answer and explanation
Answer: $4y^6$
$12 \div 3 = 4$ and $8 - 2 = 6$ .
Work out $64^{\frac{2}{3}}$ .
- $16$
- $42.7$
- $8$
- $4096$
Show answer and explanation
Answer: $16$
$\sqrt[3]{64} = 4$ , then $4^2$ .
Work out $25^{-\frac{1}{2}}$ .
- $\frac{1}{5}$
- $-5$
- $-12.5$
- $5$
Show answer and explanation
Answer: $\frac{1}{5}$
$\frac{1}{\sqrt{25}}$ .

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