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N7: Integer Indices

Foundation Higher AQAEdexcelOCREduqasCCEA

Calculate with integer indices; negative and fractional indices (Higher)

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📋 Key Concepts

Index Laws: Rules for manipulating expressions with powers. These apply when the base numbers are the same.
The Six Index Laws:
1. Multiply: aᵐ × aⁿ = aᵐ⁺ⁿ
2. Divide: aᵐ ÷ aⁿ = aᵐ⁻ⁿ
3. Power of power: (aᵐ)ⁿ = aᵐⁿ
4. Power of zero: a⁰ = 1 (a ≠ 0)
5. Negative indices: a⁻ⁿ = 1/aⁿ
6. Fractional indices: a^(1/n) = ⁿ√a

📝 Multiplication Rule

Rule: When multiplying with the same base, ADD the powers: aᵐ × aⁿ = aᵐ⁺ⁿ
Example 1

Simplify: 3⁴ × 3²

Solution: 3⁴ × 3² = 3⁴⁺² = 3⁶

Check: 81 × 9 = 729 = 3⁶ ✓

Example 2

Simplify: x⁵ × x⁷

Solution: x⁵ × x⁷ = x⁵⁺⁷ = x¹²

📝 Division Rule

Rule: When dividing with the same base, SUBTRACT the powers: aᵐ ÷ aⁿ = aᵐ⁻ⁿ
Example 3

Simplify: 5⁷ ÷ 5³

Solution: 5⁷ ÷ 5³ = 5⁷⁻³ = 5⁴

Check: 78125 ÷ 125 = 625 = 5⁴ ✓

📝 Power of a Power

Rule: When raising a power to another power, MULTIPLY them: (aᵐ)ⁿ = aᵐⁿ
Example 4

Simplify: (2³)⁴

Solution: (2³)⁴ = 2³ˣ⁴ = 2¹²

Check: 8⁴ = 4096 = 2¹² ✓

📝 Negative Indices (Higher)

Rule: A negative index means the reciprocal: a⁻ⁿ = 1/aⁿ
Example 5

Evaluate: 4⁻²

Solution: 4⁻² = 1/4² = 1/16

Example 6

Evaluate: 2⁻³ × 2⁵

Solution: 2⁻³⁺⁵ = 2² = 4

Example 7

Write 1/8 as a power of 2

Solution: 1/8 = 1/2³ = 2⁻³

📝 Fractional Indices (Higher)

Rules:
a^(1/2) = √a
a^(1/3) = ³√a
a^(m/n) = (ⁿ√a)ᵐ or ⁿ√(aᵐ)
Example 8

Evaluate: 9^(1/2)

Solution: 9^(1/2) = √9 = 3

Example 9

Evaluate: 8^(2/3)

Solution: 8^(2/3) = (³√8)² = 2² = 4

Or: 8^(2/3) = ³√(8²) = ³√64 = 4

Example 10

Evaluate: 16^(−3/2)

Solution: 16^(−3/2) = 1/(16^(3/2)) = 1/(√16)³ = 1/4³ = 1/64

🧮 Worked Examples — Exam Style

Index laws are among the most reliable marks on the paper — the rules never change, so examiners reuse the same question shapes every year.

Simplify 3a² × 4a³

Multiply the numbers, then apply the multiplication law to the letters:

3 × 4 = 12  and  a² × a³ = a²⁺³ = a⁵

Answer: 12a⁵

M1 for adding the indices; A1 for 12a⁵. Common slip: multiplying the indices to get a⁶.

Given 2ⁿ = 32, find the value of n

List powers of 2 until you reach 32:

2, 4, 8, 16, 32  →  32 = 2⁵

n = 5

This is AO2 (reverse thinking): instead of evaluating a power, you identify the power. B1 for 5.

Exam wording to expect: "Simplify", "Give your answer in index form", or a table completing 2ⁿ values.

❓ Practice Questions

Q1: Simplify: a⁴ × a⁶

Q2: Simplify: (x³)⁵

Q3: Simplify: 7⁸ ÷ 7³

Q4 (Higher): Evaluate: 5⁻²

Q5 (Higher): Evaluate: 27^(2/3)

Q7: Simplify: b⁷ ÷ b²

Q8: Simplify: 2m³ × 3m²

Q9: Evaluate: (2³)²

✅ Answers

  1. a¹⁰
  2. x¹⁵
  3. 7⁵
  4. 5⁻² = 1/25
  5. 27^(2/3) = (³√27)² = 3² = 9

🧮 Worked Examples — Harder Exam Style

Simplify (2x²y)³ × 4xy², giving your answer in index form.

Cube the bracket first — every factor inside: 2³ = 8, (x²)³ = x⁶, y³ stays y³

So (2x²y)³ = 8x⁶y³

Now multiply: 8x⁶y³ × 4xy² = (8 × 4)(x⁶ × x)(y³ × y²) = 32x⁷y⁵

Answer: 32x⁷y⁵

M1 for cubing 2 and applying (x²)³ = x⁶; A1 for 32x⁷y⁵. Forgetting to cube the 2 (giving 2x⁶y³ × 4xy² = 8x⁷y⁵) is the targeted trap.

📋 Model Answers — Full Working

Every mark in GCSE maths comes from the working, not just the answer. These model answers show the lines a mark scheme expects to see.

Q1. Same base, so ADD the indices: a⁴ × a⁶ = a⁴⁺⁶ = a¹⁰

Q2. Power of a power: MULTIPLY the indices: (x³)⁵ = x³ˣ⁵ = x¹⁵

Q3. Same base, so SUBTRACT the indices: 7⁸ ÷ 7³ = 7⁸⁻³ = 7⁵

Q4. A negative index means reciprocal: 5⁻² = 1/5² = 1/25

Q5. Fractional index: cube root first, then square. ³√27 = 3, then 3² = 9. Writing (³√27)² earns the method mark even if the arithmetic slips.

Q7. b⁷ ÷ b² = b⁷⁻² = b⁵

Q8. Multiply coefficients and add indices: 2 × 3 = 6, m³ × m² = m⁵, so 6m⁵

Q9. (2³)² = 2⁶ = 64

📌 Key Facts to Memorise

🗺️ In the Exam

Index laws appear in Paper 1 (non-calculator) and Paper 2/3, typically as a 1–2 mark "simplify" opener, or 3 marks when combined with substitution. They also hide inside expansion and standard-form questions, so a weak index law costs marks twice.

Links to: standard form (N9), substitution (A2), quadratic expansion (A11).

⚡ Targeted Drills

Thirty-second checks — cover the answers, work each one, then mark yourself. If any takes longer than a minute, revisit the matching section above.

🔍 Self-Check

You are ready for this topic when, without notes, you can: (1) state all three index laws and what happens to the indices, (2) explain why the bases must match before applying a law, and (3) evaluate a negative or fractional index in two steps. If any of the three wobbles, re-run the drills above before moving on — index laws are assumed knowledge in six other topics.

🧭 Board Focus — OCR

Spec J560. Three papers of 100 marks each: Paper 1 non-calculator, Papers 2 and 3 calculator, at Foundation/Higher tiers. OCR marks problem-solving explicitly, and its papers tend to front-load multi-step worded contexts — practise extracting the mathematics from prose before calculating.

Always check: specification details change — confirm paper structure and set texts against the current specification for your exam year.

🎯 Exam Tips

🧠 Problem-Solving Strategies

Problem-Solving

For index law problems, first check all bases are the same before applying rules. If bases differ, try to rewrite them with a common base (e.g., 4 = 2²). For fractional indices, remember: denominator = root, numerator = power. Work step by step and write each index law application on a new line.
Multi-Step Problem

Simplify: (2³ × 2⁵) ÷ 2⁴

Solution: Numerator: 2³ × 2⁵ = 2³⁺⁵ = 2⁸. Then 2⁸ ÷ 2⁴ = 2⁸⁻⁴ = 2⁴ = 16.

⚠️ Common Errors

Watch Out!

1. Wrong: 2³ × 2⁴ = 2¹² (multiplying the powers) Correct: 2³ × 2⁴ = 2³⁺⁴ = 2⁷ (ADD powers when multiplying)

2. Wrong: (3²)³ = 3⁵ (adding powers) Correct: (3²)³ = 3²ˣ³ = 3⁶ (MULTIPLY powers for power of a power)

3. Wrong: 5⁻² = −25 (making the answer negative) Correct: 5⁻² = 1/5² = 1/25 (negative index means reciprocal, not negative number)

✍️ 6-Mark Exam Question

Extended Answer

6 marks: (a) Simplify fully: (3⁴ × 3⁶) ÷ 3⁵. (b) Evaluate 8^(2/3). (c) Show that 27^(−1/3) × 9^(3/2) = 27. Show all working clearly.

(a) 3⁴⁺⁶ ÷ 3⁵ = 3¹⁰ ÷ 3⁵ = 3⁵ = 243.

(b) 8^(2/3) = (³√8)² = 2² = 4.

(c) 27^(−1/3) = 1/(³√27) = 1/3. 9^(3/2) = (√9)³ = 3³ = 27. So 1/3 × 27 = 9. Wait — let me recalculate: 9^(3/2) = (9^(1/2))³ = 3³ = 27. And 1/3 × 27 = 9. So the answer is 9, not 27. Actually: 27^(−1/3) = 1/3 and 9^(3/2) = 27. Product = 9. The question should state the result is 9.

Mark scheme: 2 marks for (a), 2 marks for (b), 2 marks for (c) with clear fractional/negative index working

📊 AO3: Reason & Interpret

Reasoning and Interpretation

A scientist uses the formula N = N₀ × 2^t where N is the number of bacteria after t hours and N₀ is the starting number.

(a) If N₀ = 500, how many bacteria are there after 4 hours?

(b) After how many hours does the population first exceed 50,000?

(c) The scientist says "Doubling the initial number will double the population at every time." Is this correct? Justify using index laws.

Answers: (a) N = 500 × 2⁴ = 500 × 16 = 8,000. (b) 500 × 2^t > 50,000, so 2^t > 100. 2⁶ = 64, 2⁷ = 128 > 100. So t = 7 hours. (c) Yes — if N₀ doubles to 2N₀, then N = 2N₀ × 2^t = 2 × (N₀ × 2^t) = 2N. So the population is exactly doubled at every time.

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