Is 25 equal to 5 or 5?

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

1MA1 · N6–N7NumberFoundation & HigherGrades 3–7

Powers & Roots

Q: Why is anything to the power 0 equal to 1?

Because a^n a^n = 1 and the division law gives a^0. Defining a^0 = 1 keeps the laws consistent.

Q: Can the base be negative with a fractional power?

Only when the root is odd, such as (-8)^1/3 = -2. GCSE questions avoid even roots of negatives.

A power is repeated multiplication written compactly: $2^{5}$ means five 2s multiplied together. Roots undo powers. Once you know a few rules for combining powers, expressions like $16^{3/4}$ or $5^{-2}$ — which look impossible at first — can be worked out in your head.

45 min Powers, Roots & Standard Form

The big picture

The index laws are not separate facts to memorise: they all follow from counting how many times a number is multiplied. Multiplying $2^{3}$ by $2^{4}$ gives seven 2s in a row, so the powers add.

Zero, negative and fractional powers are defined so that those same laws keep working. That is why $a^{0} = 1$ , why a negative power means a reciprocal, and why a power of $\tfrac12$ is a square root. Edexcel tests these without a calculator, often asking for exact values.

What you'll learn

Recall squares to $15^{2}$ , cubes of 1–5 and 10, and the corresponding roots
Use the laws of indices with numbers
Evaluate zero and negative powers
Evaluate fractional powers such as $27^{2/3}$ (Higher)

Squares, cubes and roots

A $\textbf{square number}$ is an integer multiplied by itself; a $\textbf{cube number}$ is an integer used three times: $4^{3} = 64$ .

The square root $\sqrt{81} = 9$ is the positive number that squares to 81; note that $-9$ also squares to 81, so the equation $x^{2} = 81$ has two solutions.

Cube roots can be negative: $\sqrt[3]{-125} = -5$ , because $(-5)^{3} = -125$ .

Tip — Learn squares up to $15^{2} = 225$ and cubes $1, 8, 27, 64, 125$ and $1000$ . Non-calculator questions are built around them.

The index laws

Multiplying powers of the same base: add the indices. $3^{4} \times 3^{2} = 3^{6}$ .

Dividing: subtract the indices. $7^{9} \div 7^{3} = 7^{6}$ .

Power of a power: multiply the indices. $(5^{2})^{3} = 5^{6}$ .

The laws only work when the bases are the same. $2^{3} \times 3^{2}$ cannot be combined into a single power.

a^{m} \times a^{n} = a^{m+n}

Multiply: add indices.

a^{m} \div a^{n} = a^{m-n}

Divide: subtract indices.

(a^{m})^{n} = a^{mn}

Power of a power: multiply indices.

Example 1Write

\dfrac{4^{7} \times 4^{-2}}{(4^{2})^{3}}

as a single power of 4.

1Top:

4^{7 + (-2)} = 4^{5}

2Bottom:

4^{6}

3Divide:

4^{5 - 6}

Answer

4^{-1}

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

Write $\dfrac{5^3 \times 5^{-5}}{5^{-4}}$ as a single power of 5.

Your answer

Type it like 5^3.

Show worked solution

Solution

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

Answer: $5^2$

Zero and negative powers

Using the division law, $a^{3} \div a^{3} = a^{0}$ , but anything divided by itself is 1. So $a^{0} = 1$ for any non-zero $a$ .

Similarly, $a^{0} \div a^{n} = a^{-n}$ , which is $\dfrac{1}{a^{n}}$ . A negative power means $\textbf{reciprocal}$ — it does not make the number negative.

For fractions, a negative power flips the fraction: $\left(\dfrac{2}{3}\right)^{-2} = \left(\dfrac{3}{2}\right)^{2} = \dfrac{9}{4}$ .

Example 1Work out (a)

6^{0}

(b)

2^{-3}

(c)

\left(\tfrac{1}{5}\right)^{-2}

1(a) Any non-zero number to the power 0 is 1.

2(b)

\dfrac{1}{2^{3}}

3(c) Flip to

5^{2}

Answer(a)

1

(b)

\tfrac18

(c)

25

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

Work out $\left(\frac{2}{3}\right)^{-2}$ as a fraction.

Your answer

Type a fraction like 9/4.

Show worked solution

Solution

1The negative power flips the fraction: $\left(\frac{3}{2}\right)^2$ .
2 $= \frac{9}{4}$ .

Answer: $\frac{9}{4}$

Why this works

The pattern shows why: $2^{3} = 8$ , $2^{2} = 4$ , $2^{1} = 2$ , $2^{0} = 1$ , $2^{-1} = \tfrac12$ . Each step down divides by 2, and the pattern continues smoothly past zero.

Fractional powers

A power of $\tfrac12$ is a square root, because $(a^{1/2})^{2} = a^{1} = a$ . Likewise $a^{1/3} = \sqrt[3]{a}$ and $a^{1/n} = \sqrt[n]{a}$ .

For $a^{m/n}$ , take the $n$ th root first, then raise to the power $m$ : $a^{m/n} = \left(\sqrt[n]{a}\right)^{m}$ . Rooting first keeps the numbers small.

Combine with a negative sign by taking the reciprocal as well: $8^{-2/3} = \dfrac{1}{(\sqrt[3]{8})^{2}} = \dfrac14$ .

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

Root, then power.

Example 1Work out

\left(\dfrac{16}{81}\right)^{-3/4}

1Negative power: flip to

\left(\dfrac{81}{16}\right)^{3/4}

2Fourth root:

\dfrac{3}{2}

3Cube:

\dfrac{27}{8}

Answer

\dfrac{27}{8}

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

Work out $125^{\frac{2}{3}}$ .

Your answer

Show worked solution

Solution

1 $\sqrt[3]{125} = 5$ .
2 $5^2 = 25$ .

Answer: $25$

Tip — Deal with the three parts of the index separately and in this order: negative (flip), denominator (root), numerator (power).

Think like an examiner

Show the intermediate step, such as $27^{2/3} = 3^{2}$ , not just the final answer.
Give exact answers as fractions rather than rounded decimals.
When solving for an unknown index, write both sides as powers of the same base.

Index laws

a^{0} = 1

Zero power.

a^{-n} = \tfrac{1}{a^{n}}

Negative power.

a^{1/n} = \sqrt[n]{a}

Unit fractional power.

a^{m/n} = (\sqrt[n]{a})^{m}

General fractional power.

Watch out for these

Writing

5^{-2} = -25

A negative power is a reciprocal:

\tfrac{1}{25}

Multiplying indices when multiplying powers:

3^{2} \times 3^{4} = 3^{8}

Add them:

3^{6}

Writing

7^{0} = 0

Any non-zero number to the power 0 is 1.

Evaluating

64^{2/3}

64^{2}

then cube root.

Valid but slow; cube root first:

4^{2} = 16

Stretch yourself

Find the value of $x$ such that $\dfrac{9^{x}}{27} = 3^{5} \times 81^{-1}$ .

Hint — Rewrite every term as a power of 3, then compare indices.

Reveal solution

SolutionLeft side:

\dfrac{(3^{2})^{x}}{3^{3}} = 3^{2x - 3}

. Right side:

3^{5} \times (3^{4})^{-1} = 3^{5 - 4} = 3^{1}

. Equating indices:

2x - 3 = 1

, so

x = 2

. Check:

\dfrac{81}{27} = 3

and

\dfrac{243}{81} = 3

✓.

\blacksquare

Questions students ask

Why is anything to the power 0 equal to 1?

Because

a^{n} \div a^{n} = 1

and the division law gives

a^{0}

. Defining

a^{0} = 1

keeps the laws consistent.

\sqrt{25}

equal to 5 or

\pm 5

The symbol

\sqrt{\ }

means the positive root, 5. But

x^{2} = 25

has two solutions,

x = \pm 5

Can the base be negative with a fractional power?

Only when the root is odd, such as

(-8)^{1/3} = -2

. GCSE questions avoid even roots of negatives.

Key takeaways

Know squares, cubes and their roots by heart.
Same base: multiply by adding indices, divide by subtracting, power of a power by multiplying.
$a^{0} = 1$ ; $a^{-n}$ is the reciprocal of $a^{n}$ .
$a^{m/n}$ : take the $n$ th root, then the $m$ th power.
Write everything as a power of a common base to solve index equations.

Exam-style questions

Practise it the way it is examined

3 original questions · 7 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 32 marks

(a)Write down the value of $7^0$ .[1]
Show mark scheme (a)
- B1 $1$
(b)Work out $4^{-2}$ .[1]
Show mark scheme (b)
- B1 $\frac{1}{16}$
Answer: $\frac{1}{16}$

Question 2Grade 52 marks

Work out $32^{-\frac{3}{5}}$ .[2]
Show mark scheme
- M1 $\sqrt[5]{32} = 2$ or $2^3 = 8$
- A1 $\frac{1}{8}$
Answer: $\frac{1}{8}$

Question 3Grade 73 marks

Solve $27^x = 3^{x + 4}$ .[3]
Show mark scheme
- M1 $27^x = 3^{3x}$
- M1 $3x = x + 4$
- A1 $x = 2$
Answer: $x = 2$

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