0606/12 Feb/Mar 2025, Question 9: Solving Exponential and Log Equations
Cambridge IGCSE Additional Mathematics 0606 · Paper 12 (Non-calculator) · Non-calculator · 5 marks
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The question, in short
Solve the equation log₂(x + 1) − 4 log_(x+1) 2 = 3 for x, using change of base where needed.
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Solution · 5 marks
- 1Change of base: log_(x+1) 2 = 1/log₂(x + 1)
- 2Let u = log₂(x + 1): u − 4/u = 3
- 3u² − 3u − 4 = 0
- 4(u − 4)(u + 1) = 0, so u = 4 or u = −1
- 5x + 1 = 2⁴ = 16 → x = 15; x + 1 = 2⁻¹ = 1/2 → x = −1/2
Answerx = 15, x = −1/2
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Also searched as: 0606/12/F/M/25 · 0606_m25_qp_12 · 0606 m25 12 · question 9.
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