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0606/12 May/Jun 2026, Question 9: Solving Exponential and Log Equations

Cambridge IGCSE Additional Mathematics 0606 · Paper 12 (Non-calculator) · Non-calculator · 7 marks

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The question, in short

Combine a difference involving 1/(log_x e) and ln(x − 1) into one natural logarithm, then solve 2 log₅ 2x = 1 + log₅(9x − 10).

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Part (a) · 2 marks

Express 1/(log_x e) − ln(x − 1) as one logarithm with base e.

  1. 11/(log_x e) = logₑ x = ln x
  2. 2ln x − ln(x − 1) = ln(x/(x − 1))

Answerln(x/(x − 1))

Part (b) · 5 marks

Find the value(s) of x satisfying 2 log₅ 2x = 1 + log₅(9x − 10).

  1. 12 log₅ 2x = log₅(2x)² and 1 = log₅ 5
  2. 2log₅(4x²) = log₅(5(9x − 10))
  3. 34x² = 45x − 50, so 4x² − 45x + 50 = 0
  4. 4(4x − 5)(x − 10) = 0
  5. 5x = 5/4 or x = 10 (both give 9x − 10 > 0)

Answerx = 5/4, x = 10

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Also searched as: 0606/12/M/J/26 · 0606_s26_qp_12 · 0606 s26 12 · question 9.
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