Laws of Indices

Multiply, divide and power numbers written as powers, find a missing power, and change a base like 8 into a power of 2.

In the exam for Edexcel IGCSE Maths 4MA1, Cambridge IGCSE Maths 0580, Cambridge O Level Maths 4024, Edexcel Further Pure Maths 4PM1 · 36 real past-paper questions below

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What a Power Means

A power is a short way to write the same number multiplied again and again.

5^{3} = 5 \times 5 \times 5 = 125

Rule

The big number is the base. The small number is the index (or power): it counts the copies.

So $7 \times 7 \times 7 \times 7 \times 7$ is $7^{5}$ . Count the sevens.

Common mistake

Reading $5^{3}$ as $5 \times 3 = 15$ . It is three fives multiplied: 125.

Multiplying and Dividing Powers

The 1-mark question: write it as a single power. The base must be the same number.

Formula

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

Formula

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

Here $a$ is the base, and $m$ and $n$ are the powers. Multiplying adds the powers. Dividing subtracts them.

Worked example

1
$5^{2} \times 5^{3}$ — The question. Both powers have base 5, so the law can be used.
2
$= (5 \times 5) \times (5 \times 5 \times 5)$ — Write each power out: 5² is 2 fives and 5³ is 3 fives.
3
$= 5 \times 5 \times 5 \times 5 \times 5$ — Remove the brackets. Everything is multiplied, so it is one long row of fives.
4
$= 5^{5}$ — Count them: 2 + 3 = 5 fives. That is why multiplying adds the powers.

Worked example

1
$\frac{6 ^{9} \times 6 ^{2}}{6 ^{5}}$ — The 2-mark question. Every base is 6, so only the powers change.
2
$= \frac{6 ^{9 + 2}}{6 ^{5}}$ — Do the top first. It is a multiply, so add the powers.
3
$= \frac{6 ^{11}}{6 ^{5}}$ — 9 + 2 = 11.
4
$= 6^{11 - 5}$ — Now it is a divide, so subtract the bottom power from the top power.
5
$= 6^{6}$ — 11 − 5 = 6. The answer stays as a power of 6.

Common mistake

Multiplying the bases: $5^{2} \times 5^{3} = 2 5^{5}$ . The base stays 5. Only the powers change.

Find the Missing Power

The most common question: the answer is given, and one power is hidden. Write an equation using only the powers.

Worked example

1
$\frac{2 ^{5} \times 2 ^{n}}{2 ^{3}} = 2^{10}$ — The question. Every base is 2, so you only need to work with the powers.
2
$5 + n - 3 = 10$ — Left side: multiplying adds n to 5, and dividing takes 3 away. The two sides are equal, so their powers are equal.
3
$n + 2 = 10$ — Tidy the numbers: 5 − 3 = 2.
4
$n = 8$ — Take 2 from both sides. Check: 5 + 8 − 3 = 10, so it is right.

Negative powers follow the same laws. Put a negative power in brackets so its minus sign is not lost.

Worked example

1
$4^{- 5} \times 4^{2} = 4^{m}$ — The question. One power is negative, but the law is the same: multiplying adds the powers.
2
$m = (- 5) + 2$ — Add the powers. −5 stays in a bracket so its minus stays with it.
3
$m = - 3$ — Start at −5 and go up 2. You reach −3, not 3 or 7.

Take care when the missing power is on the bottom. Dividing takes it away, so it gets a minus sign in your equation.

Worked example

1
$\frac{2 ^{9} \times 2 ^{3}}{2 ^{p}} = 2^{15}$ — The question. The missing power p is on the bottom, so it is being divided.
2
$9 + 3 - p = 15$ — Multiplying adds 9 and 3. Dividing takes p away, so write −p.
3
$12 - p = 15$ — Tidy the numbers: 9 + 3 = 12.
4
$12 = 15 + p$ — Add p to both sides, so p is no longer being taken away.
5
$p = 12 - 15 = - 3$ — Take 15 from both sides. A negative answer is fine: powers can be negative.
6
$9 + 3 - (- 3) = 15$ — Check: taking away −3 is the same as adding 3, and 12 + 3 = 15.

Common mistake

Giving $n = 10$ . The 10 is the total power, not the missing one.

Common mistake

Giving $p = 3$ . Then the left side is $2^{12 - 3} = 2^{9}$ , not $2^{15}$ . Always put your answer back in to check.

A Power of a Power

The 1-mark question: a bracket with a power outside. Multiply the powers.

Formula

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

$a$ is the base, $m$ is the power inside the bracket and $n$ is the power outside it.

Worked example

1
$(5^{2})^{3}$ — The question. The power 3 outside means three copies of what is inside.
2
$= 5^{2} \times 5^{2} \times 5^{2}$ — Write the three copies of the bracket side by side.
3
$= 5^{2 + 2 + 2}$ — Same base, multiplied, so add the powers.
4
$= 5^{6}$ — 2 + 2 + 2 is three 2s, which is 2 × 3 = 6. That is why a bracket multiplies the powers.

Worked example

1
$(2^{- 3})^{4} \times 2^{5} = 2^{k}$ — The 2-mark question. Do the bracket first, then the multiply.
2
$(2^{- 3})^{4} = 2^{(- 3) \times 4}$ — A bracket to a power: multiply the powers. Keep −3 in a bracket so the sign stays.
3
$= 2^{- 12}$ — A minus times a plus gives a minus: (−3) × 4 = −12.
4
$2^{- 12} \times 2^{5} = 2^{- 12 + 5}$ — Now multiply the powers of 2, so add the powers.
5
$k = - 7$ — Start at −12 and go up 5: −12 + 5 = −7.

Common mistake

Adding: $(5^{2})^{3} = 5^{5}$ . A bracket to a power multiplies the powers.

If the power outside the bracket is missing, the powers still multiply. So divide to find it.

Worked example

1
$(7^{3})^{k} = 7^{15}$ — The question. The power outside the bracket is missing.
2
$3 \times k = 15$ — The bracket multiplies the powers, so 3 times k must equal 15.
3
$k = 15 \div 3 = 5$ — Divide both sides by 3 to undo the multiply. Check: 3 × 5 = 15.

Key idea

The answer can be a fraction. $(3^{4})^{k} = 3^{2}$ gives $4 k = 2$ , so $k = \frac{1}{2}$ .

Change to the Same Base

The 3-mark question: the bases are different, like 8 and 4. Write each one as a power of the same prime first.

Worked example

1
$8^{2} \times 4^{3}$ — The question: write it as a single power of 2. The bases 8 and 4 are different, so the laws cannot be used yet.
2
$8 = 2^{3}$ — 8 = 2 × 2 × 2, so 8 is 2 to the power 3.
3
$4 = 2^{2}$ — 4 = 2 × 2, so 4 is 2 to the power 2.
4
$(2^{3})^{2} \times (2^{2})^{3}$ — Put them in place of 8 and 4. Keep the brackets: each outside power belongs to the whole number.
5
$= 2^{6} \times (2^{2})^{3}$ — First bracket: a power of a power multiplies, 3 × 2 = 6.
6
$= 2^{6} \times 2^{6}$ — Second bracket: 2 × 3 = 6.
7
$= 2^{12}$ — The bases match now, and it is a multiply, so add the powers: 6 + 6 = 12.

Common mistake

Writing $3 2^{5}$ . You cannot multiply different bases. Change them to the same base first.

Key idea

One number can be written in more than one base: $8 1^{2} = 9^{4} = 3^{8}$ , because 81 is $9^{2}$ and 9 is $3^{2}$ .

Find the Value: End With a Number

“Find the value” means the answer is a number, not a power. Combine the powers first, then work out the number.

Worked example

1
$4^{3} \div 4^{- 1}$ — The question. Both bases are 4, and it is a divide.
2
$= 4^{3 - (- 1)}$ — Divide, so subtract the powers. Put −1 in a bracket, because you are taking away a negative.
3
$= 4^{3 + 1}$ — Taking away −1 is the same as adding 1. Two minus signs side by side make a plus.
4
$= 4^{4}$ — 3 + 1 = 4. This is the single power.
5
$= 4 \times 4 \times 4 \times 4 = 256$ — The question asks for a value, so work out the number: 16 × 16 = 256.

Common mistake

Working out 3 − 1 = 2 and answering 16. The minus in −1 was lost: $3 - (- 1) = 4$ .

Key idea

Show the power line, like $4^{4}$ . In a 2-mark question it usually earns 1 mark on its own, even if the final number goes wrong.

Several Primes in One Number

Some questions give numbers already split into primes. The laws work on each prime on its own, because different bases never mix.

Worked example

1
$P = 2^{4} \times 3 \times 5^{2}, Q = 2 \times 3^{3}$ — The question: write $P^{2} \times Q$ in the form $2^{a} \times 3^{b} \times 5^{c}$ .
2
$P^{2} = (2^{4})^{2} \times (3)^{2} \times (5^{2})^{2}$ — Squaring P squares every prime inside it, not just the first one.
3
$= 2^{8} \times 3^{2} \times 5^{4}$ — A power of a power multiplies: 4 × 2 = 8, 1 × 2 = 2 (3 is $3^{1}$ ), and 2 × 2 = 4.
4
$P^{2} \times Q = 2^{8} \times 2 \times 3^{2} \times 3^{3} \times 5^{4}$ — Now multiply by Q. Put the same primes next to each other.
5
$= 2^{9} \times 3^{5} \times 5^{4}$ — Add the powers for each prime: 8 + 1 = 9 and 2 + 3 = 5. Q has no 5, so $5^{4}$ stays.

Common mistake

Squaring only the first prime: $2^{8} \times 3 \times 5^{2}$ . The power outside reaches every prime in P.

Which Method? Read the Question

Look at the bases first, then the sign between the powers.

You see	You do
$7 \times 7 \times 7$	Count the copies
$5^{2} \times 5^{3}$	Add the powers
$6^{11} \div 6^{5}$	Subtract the powers
$(5^{2})^{3}$	Multiply the powers
A missing power	Write an equation in the powers
A missing power on the bottom	It gets a minus: $12 - p = 15$
Bases like 8 and 4	Change to powers of 2
Several primes, like $2^{4} \times 5^{2}$	Work on each prime on its own
“Find the value”	Combine, then work out the number
A power of 0, like $7^{0}$	The answer is 1 (Zero and Negative Indices)
Letters instead of numbers	Index Laws with Letters
$2^{x} = 32$	Exponential Equations

Key idea

The laws only work when the bases match. $2^{3} \times 5^{2}$ stays as it is.

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