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Indices (laws of indices) — KCSE Mathematics

KCSE Mathematics · 100 practice questions · 4 syllabus objectives

36 easy36 medium28 hard

What You'll Learn

Key learning outcomes for this topic, aligned to the KNEC KCSE syllabus.

State and apply the laws of indices: aᵐ × aⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ, (aᵐ)ⁿ = aᵐⁿ, a⁰ = 1, a⁻ⁿ = 1/aⁿ

Simplify expressions involving positive, negative and fractional indices, including expressions with fractional bases

Solve equations involving indices by expressing both sides as powers of the same base

Indices (laws of indices)

Sample Questions

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1
easySHORT ANSWER3 marks

Define the simplification of the expression: (x^{3} × x^{-2}) ÷ x^{4}. (3 marks)

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Part (a) — 3 marks
Apply the product law: x^{3} × x^{-2} = x^{3-2} (1 mk)
Simplify to get x^{1} = x (1 mk)
Apply the division law: x ÷ x^{4} = x^{1-4} = x^{-3} (1 mk)
2
easySHORT ANSWER3 marks

Identify the simplified form of the expression: (x^{3} ÷ x^{5}) × x^{2}. (3 marks)

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Part (a) — 3 marks
Apply division law: x^{3} ÷ x^{5} = x^{3-5} = x^{-2} (1 mk)
Now multiply: x^{-2} × x^{2} = x^{-2+2} = x^{0} (1 mk)
Since x^{0} = 1, the final simplified form is 1 (1 mk)
3
easySHORT ANSWER3 marks

Define the simplified form of the expression: (x^{-3} × x^{5}) ÷ x^{2}. (3 marks)

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Part (a) — 3 marks
Apply multiplication law: x^{-3} × x^{5} = x^{(-3+5)} (1 mk)
Evaluate to get x^{2} (1 mk)
Apply division law: x^{2} ÷ x^{2} = x^{(2-2)} = x^{0} = 1 (1 mk)
4

State the simplified form of the expression: (2^{-3} × 2^{4}) ÷ 2^{-1}. (3 marks)

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