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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

❓ 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)

✅ Answers

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

🎯 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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