Why is 8 x 2^4 not equal to 8^5?

Because the product rule a^m x a^n = a^(m+n) only applies when the bases are identical. 8 and 2 are different bases. You must first rewrite 8 as a power of 2: 8 = 2^3. Then 8 x 2^4 = 2^3 x 2^4 = 2^(3+4) = 2^7 = 128. The answer 8^5 = 32768 is wrong.

Why is (2.5 x 10^4)^-3 equal to 6.4 x 10^-14 and not 2.5 x 10^-12?

Because a bracket power applies to every factor inside the bracket, including the coefficient. (2.5 x 10^4)^-3 = 2.5^-3 x (10^4)^-3. (10^4)^-3 = 10^-12. 2.5^3 = 15.625 so 2.5^-3 = 1/15.625 = 0.064 = 6.4 x 10^-2. Then 6.4 x 10^-2 x 10^-12 = 6.4 x 10^-14. The answer 2.5 x 10^-12 leaves 2.5 untouched, which is wrong.

How do you simplify (49^m)^2.5 as a power of 7?

First rewrite 49 as a power of 7: 49 = 7^2. So 49^m = (7^2)^m = 7^(2m). Then apply the power of 2.5: (7^(2m))^2.5 = 7^(2m x 2.5) = 7^(5m). Check for m = 1: 49^2.5 = 49^2 x 49^(1/2) = 2401 x 7 = 16807 = 7^5.

GCSE Maths Higher · AQA · Index laws and standard form

Rewrite the base as a prime power first

When students see $8 \times 2^4$ , the instinct is to add the indices and write $8^5$ . That answer is wrong because the product rule only applies when the bases are identical. The fix: rewrite $8 = 2^3$ first, then $2^3 \times 2^4 = 2^7$ .

A paired error appears in standard-form calculations: students raise only the power of ten to the bracket power and leave the coefficient untouched. So they write $(2.5 \times 10^4)^{-3} = 2.5 \times 10^{-12}$ instead of the correct $6.4 \times 10^{-14}$ . Both errors share the same root cause: a step is skipped before an index law is applied.

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

How to spot it in your own work

You wrote $8 \times 2^4 = 8^5$ , adding indices across two different bases without rewriting first.
You wrote $(2.5 \times 10^4)^{-3} = 2.5 \times 10^{-12}$ , applying the power only to the $10^4$ and leaving 2.5 unchanged.
You left $(49^m)^{2.5}$ as $49^{2.5m}$ without rewriting $49 = 7^2$ to express the answer as a power of 7.
You wrote $8 \times 2^6 \times 2^4 = 2^{10}$ , forgetting to include the index from rewriting $8 = 2^3$ .

An exam question that triggers it

Here is the structure of a typical AQA Higher index-laws question:

(a) Write $8 \times 2^6 \times 2^4$ as a single power of 2.

(b) Evaluate $(2.5 \times 10^4)^{-3}$ , giving your answer in standard form.

For (a): the misconception gives $8^{11}$ or some other wrong answer. The fix: rewrite $8 = 2^3$ , then $2^3 \times 2^6 \times 2^4 = 2^{13}$ .

For (b): the misconception gives $2.5 \times 10^{-12}$ . The fix: raise both factors to $-3$ : $2.5^{-3} \times 10^{-12} = 0.064 \times 10^{-12} = 6.4 \times 10^{-14}$ .

Why students fall for this

The product rule $a^m \times a^n = a^{m+n}$ is so well-drilled that students apply it to any product of powers, regardless of whether the bases match. The bases $8$ and $2$ feel related (one is a multiple of the other), so the check “are these actually the same base?” is skipped.

The standard-form error follows the same pattern. Students have practised applying powers to $10^n$ and correctly compute $(10^4)^{-3} = 10^{-12}$ , but they treat the coefficient 2.5 as a label rather than a factor. The bracket signals that the exponent governs everything inside, not just the power-of-ten term.

A third pattern is composite bases in power-of-a-power questions. Students who see $(49^m)^{2.5}$ apply the rule immediately to get $49^{2.5m}$ and stop there, not recognising that the question asks for a power of 7 and that rewriting $49 = 7^2$ is the required first step.

The fix: Rewrite to a common prime base first, then apply the index law

Step 1: check whether the bases are identical. If yes, apply the index law directly. If no, rewrite each base as a power of the common prime before proceeding.

Step 2: rewrite composite bases. $4 = 2^2$ , $8 = 2^3$ , $16 = 2^4$ , $49 = 7^2$ . Write each factor as a power of the smallest prime factor, then proceed.

Step 3: for a bracket power, raise every factor. $(a \times b)^n = a^n \times b^n$ . For $(2.5 \times 10^4)^{-3}$ this means computing $2.5^{-3}$ as well as $(10^4)^{-3}$ .

Step 4: renormalise if needed. After applying the power to the coefficient, check that the result is in the range $1 \leq a < 10$ . Here $0.064 = 6.4 \times 10^{-2}$ , so adjust the exponent accordingly.

Worked example

Simplify $8 \times 2^6 \times 2^4$ as a single power of 2.

Rewrite. $8 = 2^3$ , so
$8 \times 2^6 \times 2^4 = 2^3 \times 2^6 \times 2^4$
Add indices.
$2^{3+6+4} = 2^{13}$
Check: $2^{13} = 8192 = 8 \times 1024$ , and $2^6 \times 2^4 = 2^{10} = 1024$ . Confirmed.

Evaluate $(2.5 \times 10^4)^{-3}$ in standard form.

Raise both factors to -3.
$(2.5 \times 10^4)^{-3} = 2.5^{-3} \times (10^4)^{-3}$
Compute each part. $(10^4)^{-3} = 10^{-12}$ . $2.5^3 = 15.625$ , so $2.5^{-3} = 0.064$ .
Renormalise.
$0.064 \times 10^{-12} = 6.4 \times 10^{-2} \times 10^{-12} = 6.4 \times 10^{-14}$
Trap: $2.5 \times 10^{-12}$ leaves 2.5 untouched.

Simplify $(49^m)^{2.5}$ as a power of 7.

Rewrite. $49 = 7^2$ , so $49^m = (7^2)^m = 7^{2m}$ .
Apply the outer power.
$(7^{2m})^{2.5} = 7^{2m \times 2.5} = 7^{5m}$
Check $m=1$ : $49^{2.5} = 7^5 = 16807$ .

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Common questions

Why is $8 \times 2^4$ not equal to $8^5$ ?: The product rule $a^m \times a^n = a^{m+n}$ only applies when the bases are identical. 8 and 2 are different bases. Rewrite $8 = 2^3$ first: $2^3 \times 2^4 = 2^7 = 128$ . The trap $8^5 = 32768$ is wrong.
Why is $(2.5 \times 10^4)^{-3}$ equal to $6.4 \times 10^{-14}$ and not $2.5 \times 10^{-12}$ ?: A bracket power applies to every factor inside the bracket. $(2.5 \times 10^4)^{-3} = 2.5^{-3} \times 10^{-12}$ . Since $2.5^3 = 15.625$ and $1/15.625 = 0.064 = 6.4 \times 10^{-2}$ , the result is $6.4 \times 10^{-2} \times 10^{-12} = 6.4 \times 10^{-14}$ . The trap leaves 2.5 unchanged, which misses the step of raising the coefficient to the power.
How do you simplify $(49^m)^{2.5}$ as a power of 7?: Rewrite $49 = 7^2$ . So $49^m = (7^2)^m = 7^{2m}$ . Then $(7^{2m})^{2.5} = 7^{2m \times 2.5} = 7^{5m}$ . Check for $m=1$ : $49^{2.5} = 7^5 = 16807$ .

Related misconceptions

Negative indices mean reciprocals, not negative numbersReading the minus in 8^-5 as a sign on the result instead of recognising a^-n = 1/a^n, the companion misconception in the same Higher index-laws vertical.
Fractional indices mean roots, not divisionReading 9^(1/2) as 9 divided by 2 = 4.5 rather than the square root 3, the second misconception in the Higher index-laws vertical.

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