Mathematics • Year 9 • Unit 1 • Lesson 6
Combining Index Laws — Mixed Challenge
Pull together every index law from Unit 1: index notation (L1), evaluating powers (L2), product rule (L3), quotient rule (L4) and power-of-a-power (L5). You'll have to choose the right tool for each problem, spot a mistake in someone else's working, and tackle an open-ended challenge.
1. Mixed problems — pick the right combination
Each question uses two or three index laws from Lessons 1-5. Plan your order before you write. Show your working. 3 marks each
1.1 Simplify $\;\dfrac{(x^3)^4 \times x^2}{x^7}$.
1.2 Evaluate $\;\dfrac{(2^3)^2 \times 2^4}{2^7}$ as a single power of $2$, then as a number.
1.3 Simplify $\;(2a^4)^3 \times \dfrac{a^2}{a^8}$ fully (use power-of-a-product on the bracket — recall $(ab)^n = a^n b^n$ from Lesson 5).
1.4 Simplify $\;\dfrac{12 m^5 \times m^3}{4 m^2}$ fully.
1.5 Find $n$ if $\;\dfrac{(x^2)^n \times x^3}{x^5} = x^4$, and check your answer.
1.6 Evaluate $\;(-2)^3 \times \dfrac{(-2)^4}{(-2)^5}$. Use the product rule and quotient rule with base $-2$, then determine the sign of the final answer.
2. Find the mistake
Another student has tried to simplify $\;\dfrac{(a^3)^2 \times a^4}{a^5}$. 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 $\dfrac{(a^3)^2 \times a^4}{a^5}$:
Line 1: $(a^3)^2 = a^{3 + 2} = a^5$
Line 2: $a^5 \times a^4 = a^{5+4} = a^9$
Line 3: $\dfrac{a^9}{a^5} = a^{9-5} = a^4$
Line 4: So $\dfrac{(a^3)^2 \times a^4}{a^5} = a^4$.
(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? Compare Line 1's working to the lesson's rule for $(a^m)^n$ — are they adding or multiplying the indices? Which one should it be?3. Open-ended challenge — Build a $2^6$ expression
This question has more than one valid answer — there are several different expressions that work. 4 marks
3.1 Build two different expressions that simplify to $\;2^6$, where each expression uses at least two different index laws (chosen from: product rule, quotient rule, power-of-a-power). Use positive whole-number indices.
For each expression you build:
(i) Write it down.
(ii) Show the step-by-step working that confirms it simplifies to $2^6$.
(iii) State which two (or three) index laws you used.
Bonus: Your two expressions must use different combinations of laws (don't just submit two examples of the same pattern).
How did this worksheet feel?
What I'll revisit before next class:
1.1 — $\dfrac{(x^3)^4 \times x^2}{x^7}$
Bracket: $(x^3)^4 = x^{12}$ (power-of-a-power).
Top product: $x^{12} \times x^2 = x^{14}$ (product rule).
Quotient: $\dfrac{x^{14}}{x^7} = \mathbf{x^7}$ (quotient rule).
1.2 — $\dfrac{(2^3)^2 \times 2^4}{2^7}$
Bracket: $(2^3)^2 = 2^6$.
Top product: $2^6 \times 2^4 = 2^{10}$.
Quotient: $\dfrac{2^{10}}{2^7} = 2^3 = \mathbf{8}$.
1.3 — $(2a^4)^3 \times \dfrac{a^2}{a^8}$
Power-of-a-product on the bracket: $(2 a^4)^3 = 2^3 \times (a^4)^3 = 8 a^{12}$.
Quotient on the fraction: $\dfrac{a^2}{a^8} = a^{2-8} = a^{-6}$.
Multiply: $8 a^{12} \times a^{-6} = 8 a^{12 + (-6)} = \mathbf{8 a^6}$.
Equivalently $\dfrac{8 a^{12}}{a^6}$ if you prefer keeping the quotient unsimplified until the end.
1.4 — $\dfrac{12 m^5 \times m^3}{4 m^2}$
Top product: $m^5 \times m^3 = m^8$, so the top is $12 m^8$.
Coefficients: $12 \div 4 = 3$. Indices on $m$: $8 - 2 = 6$.
Result: $\mathbf{3 m^6}$.
1.5 — Solve $\dfrac{(x^2)^n \times x^3}{x^5} = x^4$
$(x^2)^n = x^{2n}$ (power-of-a-power).
Top product: $x^{2n} \times x^3 = x^{2n + 3}$.
Quotient: $x^{2n + 3 - 5} = x^{2n - 2}$.
Set $2n - 2 = 4$: $2n = 6$, so $\mathbf{n = 3}$.
Check: $(x^2)^3 \times x^3 / x^5 = x^6 \times x^3 / x^5 = x^9 / x^5 = x^4$ ✓.
1.6 — $(-2)^3 \times \dfrac{(-2)^4}{(-2)^5}$
Combine indices on base $-2$: $3 + 4 - 5 = 2$.
$(-2)^2 = \mathbf{+4}$ (even index of a negative base $\to$ positive).
2 — Find the mistake
(a) The mistake is on Line 1.
(b) The student has used the product rule (added the indices $3 + 2 = 5$) when they should have used the power-of-a-power rule (multiplied: $3 \times 2 = 6$). The rule for $(a^m)^n$ is $a^{mn}$ — multiply, not add. Adding indices is for $a^m \times a^n$, a different operation.
(c) Corrected working:
$(a^3)^2 = a^{3 \times 2} = a^6$
$a^6 \times a^4 = a^{6 + 4} = a^{10}$
$\dfrac{a^{10}}{a^5} = a^{10 - 5} = \mathbf{a^5}$.
This is exactly the trap from the lesson's "Spot the Trap" card: mixing add (product) with multiply (power-of-power).
3 — Open-ended (sample solution)
We need expressions that simplify to $2^6 = 64$, each using at least two index laws.
Expression 1: $(2^2)^4 \div 2^2$.
Working: Bracket: $(2^2)^4 = 2^8$ (power-of-a-power). Quotient: $2^8 \div 2^2 = 2^{8 - 2} = 2^6$ ✓.
Laws used: power-of-a-power, then quotient rule.
Expression 2: $\dfrac{2^4 \times 2^5}{2^3}$.
Working: Product on top: $2^4 \times 2^5 = 2^9$ (product rule). Quotient: $2^9 \div 2^3 = 2^{9 - 3} = 2^6$ ✓.
Laws used: product rule, then quotient rule.
Other valid examples: $(2^3)^2 \times 2^0$, $(2^5 \times 2^3) \div 2^2$, $\dfrac{(2^2)^5}{2^4}$, and many more — any combination giving a final index of $6$.
Marking: 2 marks per valid expression (one for the expression itself, one for clear step-by-step working confirming it equals $2^6$). Up to 4 in total. Award full marks for any two distinct, valid expressions using different combinations of rules.