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 · Indices, 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.

Already know this is your gap? Skip the diagnostic and jump straight into the targeted lesson.

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 (Jun22 P1 Q19 shape, non-calculator):

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 (Jun22 P1 Q19a), 8 × 2⁶ × 2⁴ as a power of 2 (Jun22 P1 Q19b), p² × p (Jun23 P3 Q10a), and y × y × y (Jun24 P2 Q2d).

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

The 10-minute diagnostic checks for this pattern (and four others) using AQA-style GCSE Higher items. Free, no signup, anonymous.

Take the diagnostic →

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.

← All GCSE Maths Higher misconceptions