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0606/13 Oct/Nov 2025, Question 9: Solving Exponential and Log Equations

Cambridge IGCSE Additional Mathematics 0606 · Paper 13 (Non-calculator) · Non-calculator · 11 marks

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The question, in short

Solve two equations: one involving log base 5 and log base 25, and one involving exponentials in y with terms divided by powers of e.

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

Part (a) · 5 marks

Solve log₅(5x − 2) − log₂₅ x = 1/2 for x.

  1. 1log₂₅ x = log₅ x/log₅ 25 = (1/2) log₅ x
  2. 2Multiply by 2: 2 log₅(5x − 2) − log₅ x = 1
  3. 3log₅((5x − 2)²/x) = 1, so (5x − 2)²/x = 5
  4. 425x² − 20x + 4 = 5x, so 25x² − 25x + 4 = 0
  5. 5(5x − 4)(5x − 1) = 0, x = 4/5 or 1/5
  6. 6x = 1/5 gives 5x − 2 < 0, so x = 4/5 only

Answerx = 4/5

Part (b) · 6 marks

Solve e^(3y−7) + 4/e³ = 5/e^(3y−1) for y.

  1. 1Multiply by e^(3y−1): e^(6y−8) + 4e^(3y−4) = 5
  2. 2Let u = e^(3y−4): u² + 4u − 5 = 0
  3. 3(u − 1)(u + 5) = 0
  4. 4u = −5 has no solution, so e^(3y−4) = 1
  5. 53y − 4 = ln 1 = 0
  6. 6y = 4/3

Answery = 4/3

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Also searched as: 0606/13/O/N/25 · 0606_w25_qp_13 · 0606 w25 13 · question 9.
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