Mathematics • Year 10 • Unit 1 • Lesson 10
Index Laws (All Five) — Mixed Challenge
Pull together every index law from the unit: multiplication and division (L8), power-of-a-power, product and quotient (L9), and zero and negative indices (L10). Choose the right tool, spot a classic mistake, and tackle an open-ended challenge.
1. Mixed problems — choose the right rule(s)
Each problem mixes laws from Lessons 8, 9 and 10. Decide which to use before you write. Show working. 3 marks each
1.1 Simplify (x⁻²)⁻⁴ and express with a positive index.
1.2 Simplify (2a⁻³)⁻², leaving with positive indices.
1.3 Evaluate (3⁻¹ + 4⁻¹). Give your answer as a single fraction.
1.4 Simplify (x³y⁻²) ÷ (x⁻¹y⁴), leaving with positive indices.
1.5 Evaluate (7² × 7⁻⁵) ÷ 7⁻³ as a single number.
1.6 Simplify ((2x⁻¹)³ × (3x²)⁻¹) ÷ x⁻⁴, leaving with positive indices.
2. Find the mistake
A student has tried to simplify 4x⁻²/2x⁻⁵, leaving with positive indices. 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 4x⁻² ÷ 2x⁻⁵:
Line 1: 4x⁻² ÷ 2x⁻⁵ = (4 ÷ 2) × (x⁻² ÷ x⁻⁵)
Line 2: = 2 × x^(−2 − (−5))
Line 3: = 2 × x^(−2 − 5)
Line 4: = 2 × x⁻⁷
Line 5: = 2/x⁷
(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? Subtracting a negative number is the same as adding the positive — i.e., −2 − (−5) = −2 + 5, not −2 − 5.3. Open-ended challenge — the same answer, three ways
This question has more than one valid answer. 4 marks
3.1 Find three different expressions, each using a different index law (Law 1, Law 2, OR the negative-index rule), that all simplify to 1/x³. At least one expression must contain a negative index that has been simplified away.
For each expression:
(i) Write it down.
(ii) Show the working that confirms it equals 1/x³ (or equivalently x⁻³).
(iii) State which law(s) you used.
Constraint: The three expressions must be different from each other and must NOT just be x⁻³, x³ ÷ x⁶, or x⁻³ × x⁰ (these are too trivial).
How did this worksheet feel?
What I'll revisit before next class:
1.1 — (x⁻²)⁻⁴
Power-of-a-power: multiply the indices. (x⁻²)⁻⁴ = x⁽⁻²⁾ˣ⁽⁻⁴⁾ = x⁸ = x⁸.
1.2 — (2a⁻³)⁻²
Power-of-a-product: 2⁻² × (a⁻³)⁻² = 1/4 × a⁶ = a⁶/4.
1.3 — 3⁻¹ + 4⁻¹
3⁻¹ = 1/3 and 4⁻¹ = 1/4. Common denominator 12: 1/3 = 4/12; 1/4 = 3/12. Sum = 7/12.
1.4 — (x³y⁻²) ÷ (x⁻¹y⁴)
x: x³⁻⁽⁻¹⁾ = x⁴.
y: y⁻²⁻⁴ = y⁻⁶ = 1/y⁶.
Combined: x⁴/y⁶.
1.5 — (7² × 7⁻⁵) ÷ 7⁻³
Combine indices: 7(2 + (−5) − (−3)) = 7(2 − 5 + 3) = 7⁰ = 1.
1.6 — ((2x⁻¹)³ × (3x²)⁻¹) ÷ x⁻⁴
(2x⁻¹)³ = 2³ × x⁻³ = 8x⁻³.
(3x²)⁻¹ = 3⁻¹ × x⁻² = (1/3) × x⁻² = x⁻²/3.
Multiply: 8x⁻³ × x⁻²/3 = (8/3) × x⁻⁵.
Divide by x⁻⁴: (8/3) × x⁻⁵⁻⁽⁻⁴⁾ = (8/3) × x⁻¹ = 8/(3x).
2 — Find the mistake
(a) The mistake is on Line 3.
(b) The student dropped the minus signs incorrectly: −2 − (−5) is equivalent to −2 + 5 = 3, NOT −2 − 5 = −7. Subtracting a negative is the same as adding the positive.
(c) Corrected working:
4x⁻² ÷ 2x⁻⁵ = (4 ÷ 2) × (x⁻² ÷ x⁻⁵)
= 2 × x⁻²⁻⁽⁻⁵⁾
= 2 × x⁻²⁺⁵
= 2 × x³
= 2x³.
Double-negative slips are the most common mistake when negative indices and Index Law 2 meet.
3 — Open-ended challenge (sample solution)
We need each expression to equal 1/x³ = x⁻³.
Expression 1 (Law 1 with a negative): x² × x⁻⁵.
Working: x² × x⁻⁵ = x²⁺⁽⁻⁵⁾ = x⁻³ = 1/x³ ✓. Law used: Index Law 1 (multiplication).
Expression 2 (Law 2): x⁴ ÷ x⁷.
Working: x⁴ ÷ x⁷ = x⁴⁻⁷ = x⁻³ = 1/x³ ✓. Law used: Index Law 2 (division).
Expression 3 (power-of-a-power with negative): (x⁻¹)³.
Working: (x⁻¹)³ = x⁽⁻¹⁾ˣ³ = x⁻³ = 1/x³ ✓. Laws used: power-of-a-power AND the negative-index rule.
Other valid choices: x⁰ ÷ x³ (zero index plus Law 2); (x³)⁻¹ (power-of-a-power); x⁻¹ × x⁻² (Law 1 with two negatives).
Marking: 1 mark per valid expression with correct working (max 3); +1 for using three genuinely different laws. Award full 4 for any three distinct, valid expressions covering at least three rules.