Mathematics • Year 9 • Unit 1 • Lesson 3
The Product Rule — Mixed Challenge
Combine every idea from Unit 1 so far: index notation (L1), evaluating powers including negatives and BIDMAS (L2), and the product rule (L3). Choose the right tool, spot a sign-and-index mistake, then take on an open-ended puzzle.
1. Mixed problems — pick the right rule
Each question uses a different combination of ideas from Lessons 1-3. Decide which rule applies before you start writing. Show your working. 3 marks each
1.1 Simplify $\;m^7 \times m^3 \times m$, then state the base and index of your answer.
1.2 Evaluate $\;2^4 \times 2^3$ as a single power of $2$, then as a number.
1.3 Simplify $\;2a^3 \times 5a^2$ fully (handle coefficients and indices separately).
1.4 Simplify $\;5^3 \times 4^2 \times 5^4$ by grouping same-base powers first.
1.5 Find $n$ if $\;x^6 \times x^n = x^{14}$, and check by substituting your value.
1.6 Evaluate $\;(-2)^3 \times (-2)^4$ in two ways: (i) by the product rule (treat $-2$ as the base), then (ii) by evaluating each power separately and multiplying. Both methods should match.
2. Find the mistake
Another student has tried to simplify $\;a^3 \times a^5$. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — simplify $a^3 \times a^5$:
Line 1: Same base $a$, so the product rule applies.
Line 2: Combine the indices: $3 \times 5 = 15$.
Line 3: Base stays the same: $a$.
Line 4: So $a^3 \times a^5 = a^{15}$.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected final answer.
Stuck? Expand: $a^3 \times a^5 = (a \cdot a \cdot a)(a \cdot a \cdot a \cdot a \cdot a) = a^?$. Count the $a$s — is it $15$ or something else?3. Open-ended challenge — Three pairs that multiply to $2^{10}$
This question has more than one valid answer — there are several different pairs that work. 4 marks
3.1 Find three different pairs of powers of $2$, of the form $2^m \times 2^n$ (with $m$ and $n$ positive whole numbers, both at least $1$, and $m \ne n$ in at least two of the pairs), that all multiply to give $\;2^{10}$.
For each pair you find:
(i) Write it down.
(ii) Show the working that confirms it equals $2^{10}$.
(iii) State which rule you used.
Bonus: Your pairs must not include $2^0 \times 2^{10}$ or $2^{10} \times 2^0$.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — $m^7 \times m^3 \times m$
$m = m^1$. Add: $7 + 3 + 1 = 11$. So $\mathbf{m^{11}}$. Base $= m$, index $= 11$.
1.2 — $2^4 \times 2^3$
$2^{4+3} = \mathbf{2^7 = 128}$.
1.3 — $2a^3 \times 5a^2$
Coefficients: $2 \times 5 = 10$. Indices on $a$: $3 + 2 = 5$. So $\mathbf{10 a^5}$.
1.4 — $5^3 \times 4^2 \times 5^4$
Group same-base: $(5^3 \times 5^4) \times 4^2 = 5^{3+4} \times 4^2 = \mathbf{5^7 \times 4^2}$.
Cannot combine $5^7$ and $4^2$ further (different bases). As a number: $5^7 = 78{,}125$ and $4^2 = 16$, so $78{,}125 \times 16 = 1{,}250{,}000$.
1.5 — Solve $x^6 \times x^n = x^{14}$
Product rule: $x^{6 + n} = x^{14}$, so $6 + n = 14$, giving $\mathbf{n = 8}$.
Check: $x^6 \times x^8 = x^{6+8} = x^{14}$ ✓.
1.6 — $(-2)^3 \times (-2)^4$ — two methods
(i) Product rule with base $-2$: $(-2)^{3+4} = (-2)^7$. Odd index keeps the sign: $(-2)^7 = -128$.
(ii) Separate evaluation: $(-2)^3 = -8$; $(-2)^4 = +16$. Multiply: $(-8) \times 16 = -128$.
Both give $\mathbf{-128}$ ✓.
2 — Find the mistake
(a) The mistake is on Line 2.
(b) The student has multiplied the indices ($3 \times 5 = 15$) instead of adding them. The product rule $a^m \times a^n = a^{m+n}$ tells you to ADD when MULTIPLYING same-base powers — because $a^3 \times a^5$ is $3 + 5 = 8$ copies of $a$ multiplied, not $3 \times 5 = 15$.
(c) Corrected working:
$a^3 \times a^5$ — same base $a$, product rule applies.
Add the indices: $3 + 5 = 8$.
Base stays the same: $a$.
So $a^3 \times a^5 = \mathbf{a^8}$.
Quick verification by expansion: $(a \cdot a \cdot a)(a \cdot a \cdot a \cdot a \cdot a) = a^8$ — eight $a$s in total.
3 — Open-ended (sample solution)
We need $2^m \times 2^n = 2^{10}$, so $m + n = 10$ with $m, n \ge 1$.
Pair 1: $2^1 \times 2^9$. Working: $2^{1 + 9} = 2^{10}$ ✓. Rule: product rule.
Pair 2: $2^3 \times 2^7$. Working: $2^{3 + 7} = 2^{10}$ ✓. Rule: product rule.
Pair 3: $2^5 \times 2^5$. Working: $2^{5 + 5} = 2^{10}$ ✓. Rule: product rule.
Other valid pairs: $2^2 \times 2^8$, $2^4 \times 2^6$ — any $(m, n)$ with $m + n = 10$ and both $\ge 1$.
Marking: 1 mark per correct pair with working — up to 3. 1 extra mark for clearly naming the product rule. Award full marks for any three distinct valid pairs other than $(0, 10)$ or $(10, 0)$.