What is 2³ × 2⁴ as a single power?

2⁷. When you multiply powers of the same base, keep the base and add the indices: 3 + 4 = 7, so 2³ × 2⁴ = 2⁷ = 128. A power is repeated multiplication, so 2³ × 2⁴ is seven 2s multiplied together — the base stays 2. The common wrong answer 4⁷ multiplies the bases (2 × 2 = 4) and gives 16384, far too big.

What is 3¹² ÷ 3⁷ as a whole number?

3⁵ = 243. Dividing powers of the same base keeps the base and subtracts the indices: 12 − 7 = 5, so 3¹² ÷ 3⁷ = 3⁵ = 243. The common trap cancels the base to write 1⁵ = 1, but dividing removes seven 3s from twelve 3s, leaving five 3s — not nothing.

How do you simplify p² × p?

p³. A lone letter is to the power 1, so p² × p = p² × p¹, and you add the indices: 2 + 1 = 3, giving p³. The common wrong answer 2p² folds the extra p into the coefficient instead of adding it to the index.

GCSE Maths Foundation · AQA · Powers & standard form

Index laws: keep the base, add the indices

Index-law questions trip students who operate on the base. Asked to write $2^3 \times 2^4$ as a single power, they answer 4⁷ — multiplying the two bases, $2 \times 2 = 4$ — instead of keeping the base and adding the indices. But a power is repeated multiplication: $2^3 \times 2^4$ is seven 2s multiplied together, so the base stays 2 and $2^3 \times 2^4 = 2^7$ .

The thirty-second fix: keep the base. Add the indices to multiply, subtract them to divide, and convert any number to the right base first. The base never turns into a bigger number and never cancels to 1.

Ready to fix this? The Powers & standard form lesson works through this misconception and the others in Powers & standard form, one altitude at a time.

How to spot it in your own work

You multiplied the bases — writing $2^3 \times 2^4 = 4^7$ instead of $2^7$ .
You cancelled the base when dividing — writing $3^{12} \div 3^7 = 1^5 = 1$ instead of $3^5$ .
You multiplied the indices instead of adding them — $3 \times 4 = 12$ , giving $2^{12}$ , instead of $3 + 4 = 7$ .
You folded a coefficient into the index — writing $p^2 \times p = 2p^2$ instead of $p^3$ .
You left a coefficient unconverted — writing $8 \times 2^4$ as it stands instead of $2^3 \times 2^4 = 2^7$ .

An exam question that triggers it

Here is the canonical AQA Foundation trigger, the shape of Jun22 P1 Q19:

Write 2³ × 2⁴ as a single power of 2.

The misconception answer is 4⁷ — multiplying the bases, $2 \times 2 = 4$ . But the base stays 2, because $2^3 \times 2^4$ is seven 2s multiplied together.

Keep the base and add the indices: $2^3 \times 2^4 = 2^{3+4} = 2^7$ .

Why students fall for this

A power like $2^4$ packs two numbers together — a base and an index — and under pressure students reach for the most visible operation, the multiplication sign, and apply it to the base. So $2^3 \times 2^4$ becomes 4⁷. But the base records what is being multiplied, and the index records how many. Multiplying powers of the same base just stacks the factors, so you count them — you add the indices, and the base is unchanged.

On division the same belief cancels the base. Students see $3^{12} \div 3^7$ , cancel the 3s, and write $1^5 = 1$ . But dividing removes factors: twelve 3s lose seven 3s and five remain, so $3^{12} \div 3^7 = 3^5$ . Cancelling to 1 would only be right if the indices were equal.

The belief has two further faces. The coefficient stays unconverted — $8 \times 2^4$ is left alone instead of writing $8 = 2^3$ so it becomes $2^3 \times 2^4 = 2^7$ . And in algebra a coefficient is folded into the index: $p^2 \times p$ becomes 2p² instead of p³. AQA Foundation papers exploit every face — 3¹² ÷ 3⁷ as a whole number, 8 × 2⁶ × 2⁴ as a power of 2, p² × p, and y × y × y.

The fix: Keep the base; add the indices to multiply, subtract them to divide, and convert any number to the base first

Multiplying: keep the base, add the indices. $2^3 \times 2^4 = 2^{3+4} = 2^7$ — seven 2s multiplied together. The base never becomes 4.

Dividing: keep the base, subtract the indices. $3^{12} \div 3^7 = 3^{12-7} = 3^5$ — never cancel the base to 1.

Convert any number to the base first. $8 = 2^3$ , so $8 \times 2^4 = 2^3 \times 2^4 = 2^7$ . And a lone letter is to the power 1: $p^2 \times p = p^{2+1} = p^3$ .

Worked example

Write 8 × 2⁶ × 2⁴ as a single power of 2.

Convert the coefficient to the base. $8 = 2 \times 2 \times 2 = 2^3$ .
Rewrite as a product of powers of 2.
$8 \times 2^6 \times 2^4 = 2^3 \times 2^6 \times 2^4$
Keep the base and add the indices.
$2^3 \times 2^6 \times 2^4 = 2^{3+6+4} = 2^{13}$

The same rule, in reverse, divides: $3^{12} \div 3^7 = 3^{12-7} = 3^5 = 243$ — not $1^5 = 1$ . Keep the base, work the indices.

Find out if this is costing you marks

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Common questions

Why don’t I multiply the bases in 2³ × 2⁴?: Because a power is repeated multiplication, and $2^3 \times 2^4$ is seven 2s multiplied together — every factor is still a 2. So the base stays 2 and you count the factors by adding the indices: $2^3 \times 2^4 = 2^7$ . Multiplying the bases gives 4⁷ = 16384, far too big.
Why isn’t 3¹² ÷ 3⁷ equal to 1?: Because dividing removes factors, it does not cancel the whole base. Twelve 3s lose seven 3s and five 3s remain, so $3^{12} \div 3^7 = 3^{12-7} = 3^5 = 243$ . Cancelling to $1^5 = 1$ would only be right if the two indices were equal.
Do I add or multiply the indices?: Add them when you multiply powers, subtract them when you divide — never multiply the indices. $2^3 \times 2^4 = 2^{3+4} = 2^7$ , not $2^{3 \times 4} = 2^{12}$ . Multiplying the indices is a different rule (for a power of a power, like $(2^3)^4$ ).
What is p² × p?: p³. A lone p is $p^1$ , so $p^2 \times p = p^{2+1} = p^3$ . The wrong answer 2p² folds the extra p into the coefficient; you must add it to the index instead.

Related misconceptions

Powers are repeated multiplication, not base × indexThe idea the index laws rest on — 2³ × 2⁴ = 2⁷ only makes sense once a power means counting copies of the base.
Squaring is not doublingThe same base-vs-index slip in the n = 2 case — 19² is 19 × 19 = 361, not 19 × 2.
Speed, distance, time & rates: convert the time before you divideA neighbouring Foundation procedure — convert to the right unit before you operate, just as you convert a coefficient to the base before applying an index law.

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