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WMA11/01 Jan 2025, Question 4: Powers with Letters in the Exponent

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 1 (WMA11) · Calculator · 9 marks

The question, in short

Using y = aˣ with a positive, rewrite two expressions in terms of y and a, then transform and solve an equation involving 9ᵗ via p = 9ᵗ.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (ia) · 1 mark

Write a³ˣ⁺¹ in its simplest form using y and a.

  1. 1a³ˣ⁺¹ = a × (aˣ)³
  2. 2= ay³

Answeray³

Part (ib) · 3 marks

Simplify 5 divided by (3a¹⁻ˣ)², giving the result in terms of y and a.

  1. 15/(3a¹⁻ˣ)⁻² = 5 × (3a¹⁻ˣ)²
  2. 2= 5 × 9 × a²⁻²ˣ = 45a² × a⁻²ˣ
  3. 3a⁻²ˣ = 1/(aˣ)² = 1/y²
  4. 4= 45a²/y²

Answer45a²/y²

Part (iia) · 2 marks

With p = 9ᵗ, show that 3(3⁴ᵗ⁺² + 1) = 82 × 9ᵗ becomes 27p² − 82p + 3 = 0.

  1. 13⁴ᵗ⁺² = 3² × 3⁴ᵗ = 9 × (3²ᵗ)² = 9 × (9ᵗ)² = 9p²
  2. 2So 3(9p² + 1) = 82p
  3. 327p² + 3 = 82p
  4. 427p² − 82p + 3 = 0

AnswerShown: 3⁴ᵗ⁺² = 9 × (3²ᵗ)² = 9p², so 3(9p² + 1) = 82p ⇒ 27p² − 82p + 3 = 0

Part (iib) · 3 marks

Using the quadratic from part (a), find the value(s) of t satisfying the original equation.

  1. 127p² − 82p + 3 = (27p − 1)(p − 3) = 0
  2. 2p = 3 or p = 1/27
  3. 39ᵗ = 3 = 9^(1/2), so t = 1/2
  4. 49ᵗ = 1/27 = 3⁻³ = 9^(−3/2), so t = −3/2

Answert = 1/2, t = −3/2

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