0606/13 May/Jun 2025, Question 4: Solving Exponential and Log Equations
Cambridge IGCSE Additional Mathematics 0606 · Paper 13 (Non-calculator) · Non-calculator · 5 marks
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The question, in short
Solve 2/(logₓ 10) − lg(x + 4) = lg 2 for x > 0, showing the value(s) of x.
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Solution · 5 marks
- 11/(logₓ 10) = lg x, so 2/(logₓ 10) = 2lg x
- 22lg x − lg(x + 4) = lg 2
- 3lg(x²/(x + 4)) = lg 2
- 4x² = 2(x + 4), so x² − 2x − 8 = 0
- 5(x − 4)(x + 2) = 0
- 6x > 0, so x = 4 only
Answerx = 4
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Also searched as: 0606/13/M/J/25 · 0606_s25_qp_13 · 0606 s25 13 · question 4.
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