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4PM1/02R Jan 2023, Question 10: Solving Exponential and Log Equations

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2R · Calculator · 13 marks

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The question, in short

Show an index identity for 9^(3y)/243, then solve simultaneous equations linking powers of 3 and a log equation in x and y.

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

Part (a) · 4 marks

Show that 9^(3y) divided by 243 simplifies to 3^(6y − 5).

  1. 11/243 = 3⁻⁵
  2. 29^(3y) = 3^(6y)
  3. 39^(3y)/243 = 3^(6y) × 3⁻⁵ = 3^(6y − 5)

Answer9^(3y)/243 = 3^(6y − 5) shown

Part (b) · 9 marks

Solve the simultaneous equations 9^(3y)/243 = 27^(x − 2) and log₁₀ √(6xy) = log₄ 2.

  1. 127^(x − 2) = 3^(3x − 6), so 6y − 5 = 3x − 6 → 3x − 6y − 1 = 0
  2. 2log₄ 2 = 1/2
  3. 3log₁₀ √(6xy) = ½ log₁₀(6xy) = ½ → log₁₀(6xy) = 1 → 6xy = 10
  4. 46y = 10/x: 3x − 10/x − 1 = 0 → 3x² − x − 10 = 0
  5. 5(3x + 5)(x − 2) = 0 → x = 2 or x = −5/3
  6. 6x = 2: y = 5/6; x = −5/3: y = −1

Answerx = 2, y = 5/6 or x = −5/3, y = −1

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