Mathematics • Year 9 • Unit 1 • Lesson 3
The Product Rule
Build fluency with $a^m \times a^n = a^{m+n}$: when multiplying powers with the SAME base, ADD the indices. One step at a time, from a fully worked example through guided practice to independent problems.
1. I do — fully worked example
Read every line. Each step has a short reason on the right.
Problem. Simplify $\;5^3 \times 5^4$. Leave your answer in index form.
Step 1 — Check the bases.
Both bases are $5$.
Reason: the product rule $a^m \times a^n = a^{m+n}$ only applies when the bases are the SAME.
Step 2 — Add the indices.
$3 + 4 = 7$
Reason: $5^3$ is three $5$s; $5^4$ is four $5$s; multiplied together that's $3 + 4 = 7$ copies of $5$.
Step 3 — Write the result with the same base.
$5^3 \times 5^4 = 5^7$
Reason: the base $5$ does NOT change — only the index changes.
Step 4 — (Optional) verify by expanding.
$(5 \cdot 5 \cdot 5) \times (5 \cdot 5 \cdot 5 \cdot 5) = \underbrace{5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5}_{7 \text{ copies}} = 5^7$ ✓
Answer: $\mathbf{5^7}$ (which equals $78{,}125$ if you want a number).
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank line. 4 marks
Problem. Simplify $\;a^2 \times a^6 \times a$.
Step 1 — Check the bases: all three bases are __________ .
Step 2 — Spot the hidden index: $a = a^{\_\_}$.
Step 3 — Add all the indices:
$2 + 6 + \_\_ = \_\_\_$
Step 4 — Write the result with the same base:
$a^2 \times a^6 \times a = a^{\_\_\_}$
Answer: $a^{\_\_}$.
3. You do — independent practice
Show your working in the space under each problem. The first four are foundation (single rule). The middle two are standard (combine two ideas). The last two are extension (push your thinking).
Foundation — single rule
3.1 Simplify $\;3^4 \times 3^2$. 1 mark
3.2 Simplify $\;x^5 \times x^3$. 1 mark
3.3 Simplify $\;7 \times 7^5$. 1 mark
3.4 Evaluate $\;2^4 \times 2^3$ as a single power of $2$, then as a number. 1 mark
Standard — combine two ideas
3.5 Simplify $\;y^4 \times y^2 \times y^3$. 2 marks
3.6 Simplify $\;2^3 \times 5^2 \times 2^4$ by grouping same-base powers first. 2 marks
Extension — push your thinking
3.7 Simplify $\;3a^2 \times 4a^5$ fully, including the coefficients. 3 marks
3.8 Find the value of $n$ if $\;x^4 \times x^n = x^{11}$. Explain how you used the product rule. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded $a^2 \times a^6 \times a$)
Step 1: all three bases are $a$.
Step 2: $a = a^{\mathbf{1}}$ (hidden index).
Step 3: $2 + 6 + \mathbf{1} = \mathbf{9}$.
Step 4: $a^2 \times a^6 \times a = a^{\mathbf{9}}$.
Answer: $\mathbf{a^9}$.
3.1 — $3^4 \times 3^2$
Same base, add indices: $4 + 2 = 6$. So $3^4 \times 3^2 = \mathbf{3^6}$.
3.2 — $x^5 \times x^3$
$x^{5+3} = \mathbf{x^8}$.
3.3 — $7 \times 7^5$
$7 = 7^1$ (hidden index). $7^1 \times 7^5 = 7^{1+5} = \mathbf{7^6}$.
3.4 — $2^4 \times 2^3$
$2^{4+3} = \mathbf{2^7} = 128$.
3.5 — $y^4 \times y^2 \times y^3$
Same base, add all three indices: $4 + 2 + 3 = 9$. So $\mathbf{y^9}$.
3.6 — $2^3 \times 5^2 \times 2^4$
Group same-base: $(2^3 \times 2^4) \times 5^2 = 2^{3+4} \times 5^2 = \mathbf{2^7 \times 5^2}$.
Cannot combine $2^7$ and $5^2$ further (different bases). As a number: $128 \times 25 = 3200$.
3.7 — $3a^2 \times 4a^5$
Multiply coefficients separately, add indices separately:
Coefficients: $3 \times 4 = 12$.
Indices on $a$: $2 + 5 = 7$.
Result: $\mathbf{12 a^7}$. (Common slip: writing $7a^7$ — forgetting that the coefficients multiply, not add.)
3.8 — Solve $x^4 \times x^n = x^{11}$
By the product rule, $x^4 \times x^n = x^{4+n}$. Match the indices: $4 + n = 11$, so $\mathbf{n = 7}$.
The rule says we add the indices when multiplying same-base powers, so the resulting index is $4 + n$. Set that equal to the given index $11$ and solve.