Mathematics • Year 9 • Unit 1 • Lesson 10
Index Laws — Mixed Challenge
Pull together every index rule from Unit 1 (L1-L10): product, quotient, power, zero, negative — all on numerical bases. Choose the right rule, spot a tricky mistake, and tackle an open-ended challenge.
1. Mixed problems — choose the right rule
Each question uses a different combination of rules from Lessons 1-10. Decide which rule applies before you start. Show working; finish with a positive index. 3 marks each
1.1 Evaluate $\dfrac{2^6 \times 2^{-2}}{2^3}$.
1.2 Evaluate $4^3 \times 4^{-3}$ using index laws (not by computing $4^3$ first).
1.3 Evaluate $(2^3)^{-1} \times 2^5$.
1.4 Evaluate $\dfrac{3^{-2} \times 3^4}{3}$.
1.5 Find $n$ if $\dfrac{6^n \times 6^{-2}}{6^3} = 6^4$.
1.6 Evaluate $\dfrac{6^2 \times 6^{-3}}{6^{-2}}$, giving your final answer as a single power and as a number.
2. Find the mistake
Another student has tried to simplify $\dfrac{2^3 \times 2^{-5}}{2^{-1}}$. 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 $\dfrac{2^3 \times 2^{-5}}{2^{-1}}$:
Line 1: Top: $2^3 \times 2^{-5} = 2^{3 + (-5)} = 2^{-2}$
Line 2: $\dfrac{2^{-2}}{2^{-1}} = 2^{-2 - 1} = 2^{-3}$
Line 3: $2^{-3} = \dfrac{1}{2^3} = \dfrac{1}{8}$
Line 4: Final answer $= \dfrac{1}{8}$.
(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? "Subtract the bottom index" means subtract $-1$, not subtract $1$. Subtracting a negative ADDS.3. Open-ended challenge — design your own
This question has more than one valid answer. 4 marks
3.1 Design one expression on a single numerical base that uses ALL five index rules (product, quotient, power, zero, negative) in one chain and evaluates to a final answer of $\mathbf{2}$. The expression must be non-trivial — at least one term must use a negative index and at least one bracket must have a power outside it.
For your expression:
(i) Write it down clearly.
(ii) Identify each of the five rules you've used, in the order you'll apply them.
(iii) Show step-by-step working that lands at the final answer $2$.
Hint to start: $2 = 2^1$. So you need the overall index to simplify to $1$. Build a chain of indices that adds and subtracts to $1$.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — $\dfrac{2^6 \times 2^{-2}}{2^3}$
Top: $2^{6 + (-2)} = 2^4$. Whole: $\dfrac{2^4}{2^3} = 2^{4-3} = 2^1 = \mathbf{2}$.
1.2 — $4^3 \times 4^{-3}$ via index laws
Product rule: $4^{3 + (-3)} = 4^0$. Zero-index rule: $4^0 = \mathbf{1}$.
1.3 — $(2^3)^{-1} \times 2^5$
Power: $(2^3)^{-1} = 2^{3 \times (-1)} = 2^{-3}$. Product: $2^{-3} \times 2^5 = 2^{-3+5} = 2^2 = \mathbf{4}$.
1.4 — $\dfrac{3^{-2} \times 3^4}{3}$
Top: $3^{-2+4} = 3^2$. (Bottom is $3 = 3^1$.) Whole: $\dfrac{3^2}{3^1} = 3^{2-1} = 3^1 = \mathbf{3}$.
1.5 — Solve $\dfrac{6^n \times 6^{-2}}{6^3} = 6^4$
Left side: $6^{n + (-2) - 3} = 6^{n-5}$. Match indices: $n - 5 = 4$, so $\mathbf{n = 9}$.
1.6 — $\dfrac{6^2 \times 6^{-3}}{6^{-2}}$
Top: $6^{2 + (-3)} = 6^{-1}$. Whole: $\dfrac{6^{-1}}{6^{-2}} = 6^{-1 - (-2)} = 6^{-1 + 2} = 6^1 = \mathbf{6}$.
Subtracting a negative adds — easy place to slip.
2 — Find the mistake
(a) The mistake is on Line 2.
(b) When dividing by $2^{-1}$, the quotient rule says to SUBTRACT the bottom index, which is $-1$. So the new index is $-2 - (-1) = -2 + 1 = -1$, not $-3$. The student subtracted $+1$ instead of $-1$ — they ignored the minus sign on the bottom index.
(c) Corrected working:
Top: $2^3 \times 2^{-5} = 2^{3 + (-5)} = 2^{-2}$.
Quotient: $\dfrac{2^{-2}}{2^{-1}} = 2^{-2 - (-1)} = 2^{-2 + 1} = 2^{-1}$.
Positive index: $2^{-1} = \dfrac{1}{2}$.
Final answer $= \mathbf{\dfrac{1}{2}}$.
This is exactly the trap flagged in the lesson — "subtract a negative adds".
3 — Open-ended challenge (sample solution)
Expression: $\dfrac{(2^2)^3 \times 2^{-4} \times 2^0}{2^{-1} \times 2^2}$.
Rules used (in order):
1. Power-of-a-power on the bracket: $(2^2)^3 = 2^{2 \times 3} = 2^6$.
2. Zero index: $2^0 = 1$.
3. Product rule on numerator: $2^6 \times 2^{-4} \times 1 = 2^{6 + (-4)} = 2^2$.
4. Product rule on denominator: $2^{-1} \times 2^2 = 2^{-1+2} = 2^1$.
5. Quotient rule across the fraction: $\dfrac{2^2}{2^1} = 2^{2-1} = 2^1 = \mathbf{2}$ ✓.
Other valid approaches: any non-trivial expression on a single base that collapses to $a^1$ — for example $\dfrac{2^5 \times 2^{-3}}{(2^0)^7 \times 2^{-(-1)+0}}$ or $(2^3)^0 \times 2^1 \times 2^0 \div 2^0$ are also valid. Award full marks for any expression that uses all five rules in working AND lands at $2$.
Marking: 1 mark per rule clearly used and labelled (max 4 for the five labels — generous overlap allowed); but ALSO require the final answer to be $2$ with correct working. Cap at 4 marks total.