Mathematics • Year 10 • Unit 1 • Lesson 9
Index Laws — Mixed Challenge
Pull together every index law from Lesson 8 (multiplication, division) and Lesson 9 (power of a power, product, quotient). Choose the right tool, spot a classic mistake, and tackle an open-ended challenge.
1. Mixed problems — choose the right rule
Each question uses a different combination of the index laws from Lessons 8-9. Decide which rule applies before you write. Show working. 3 marks each
1.1 Simplify (x⁴)³ ÷ x⁵. Leave in index form.
1.2 Simplify (2a³)⁴ × a². Show every step.
1.3 Simplify (3m²/n)³, stating any restriction on n.
1.4 Show that (−2)⁴ and −2⁴ give different answers, and explain why in one sentence. (Brackets matter.)
1.5 Find n if (5³)ⁿ = 5¹⁵, and check by also evaluating (5³)ⁿ directly.
1.6 Simplify ((2x²)³ × (3x)²) ÷ (6x⁴).
2. Find the mistake
A student has tried to simplify (2a³)⁴. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why, then re-do correctly. 3 marks
Student's working — simplify (2a³)⁴:
Line 1: (2a³)⁴ = 2⁴ × (a³)⁴
Line 2: 2⁴ = 16
Line 3: (a³)⁴ = a^(3+4) = a⁷
Line 4: So (2a³)⁴ = 16a⁷.
(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 3's working to the lesson's rule for (aᵐ)ⁿ — are they adding or multiplying the indices?3. Open-ended challenge — the same answer, two ways
This question has more than one valid answer. 4 marks
3.1 Find two different expressions of the form (aᵐ)ⁿ — where a, m and n are positive whole numbers and a > 1 — that both simplify to 2¹².
For each expression:
(i) Write it down.
(ii) Show the working that confirms it equals 2¹².
(iii) State which index law(s) you used.
Constraint: Your two expressions must be different from each other and must NOT be (2¹²)¹ or (2¹)¹².
How did this worksheet feel?
What I'll revisit before next class:
1.1 — (x⁴)³ ÷ x⁵
Numerator: (x⁴)³ = x⁴ˣ³ = x¹² (power-of-a-power).
Whole expression: x¹² ÷ x⁵ = x¹²⁻⁵ = x⁷ (Law 2).
1.2 — (2a³)⁴ × a²
(2a³)⁴ = 2⁴ × (a³)⁴ = 16 × a¹² = 16a¹² (power-of-a-product, then power-of-a-power).
× a²: 16a¹² × a² = 16 × a¹²⁺² = 16a¹⁴ (Law 1).
1.3 — (3m²/n)³
Cube top and bottom: (3m²)³ / n³ = (3³ × (m²)³) / n³ = (27 × m⁶) / n³ = 27m⁶/n³, with n ≠ 0.
1.4 — (−2)⁴ vs −2⁴
(−2)⁴ = (−2) × (−2) × (−2) × (−2) = 16 (four negatives multiplied gives a positive).
−2⁴ = −(2 × 2 × 2 × 2) = −16 (the power only applies to the 2; the minus is "out the front").
Brackets show what the index applies to. In (−2)⁴ the index applies to the whole "−2". In −2⁴ it applies only to 2.
1.5 — Solve (5³)ⁿ = 5¹⁵
By power-of-a-power, (5³)ⁿ = 5³ⁿ. Match indices: 3n = 15, so n = 5.
Check: (5³)⁵ = 5³ˣ⁵ = 5¹⁵ ✓.
1.6 — ((2x²)³ × (3x)²) ÷ (6x⁴)
Step 1 — power-of-a-product on each bracket:
(2x²)³ = 2³ × (x²)³ = 8x⁶.
(3x)² = 3² × x² = 9x².
Step 2 — multiply the numerator:
8x⁶ × 9x² = 72 × x⁶⁺² = 72x⁸.
Step 3 — divide by 6x⁴:
72x⁸ ÷ 6x⁴ = (72/6) × x⁸⁻⁴ = 12x⁴.
2 — Find the mistake
(a) The mistake is on Line 3.
(b) The student added the indices (3 + 4 = 7) instead of multiplying them. The rule for (aᵐ)ⁿ is power-of-a-power, which multiplies: (a³)⁴ = a³ˣ⁴ = a¹². Adding indices is Index Law 1, which applies to aᵐ × aⁿ — a different situation.
(c) Corrected working:
(2a³)⁴ = 2⁴ × (a³)⁴
= 16 × a³ˣ⁴
= 16 × a¹²
= 16a¹².
This is precisely the trap flagged in the lesson's "Common Pitfalls" card: confusing Law 1 (add) with power-of-a-power (multiply).
3 — Open-ended challenge (sample solution)
We need (aᵐ)ⁿ = 2¹², so we need a to be a power of 2 and the resulting index mn (after applying any rewriting) to equal 12. Easiest: keep a = 2 and pick m, n with m × n = 12.
Expression 1: (2²)⁶.
Working: (2²)⁶ = 2²ˣ⁶ = 2¹² ✓.
Rule used: power-of-a-power.
Expression 2: (2³)⁴.
Working: (2³)⁴ = 2³ˣ⁴ = 2¹² ✓.
Rule used: power-of-a-power.
Other valid approaches: (2⁴)³, (2⁶)² — any pair (m, n) with m × n = 12 and m, n ≥ 2. Students who notice 4 = 2² can also write (4³)² = ((2²)³)² = 2¹², using power-of-a-power twice.
Marking: 2 marks per valid expression (one for the expression itself, one for the working). Up to 4 in total. Award full marks for any two distinct valid expressions other than the trivial (2¹²)¹ or (2¹)¹².