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4PM1/01R Jun 2024, Question 7: Solving Exponential and Log Equations

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1R · Calculator · 9 marks

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The question, in short

Sketch the curve y = −log₄(x + 4) on the given axes, marking axis crossings and asymptote, then solve log₍ₓ₊₄₎ 256 − log₄(x + 4) = 0.

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

Part (a) · 4 marks

On the blank axes, draw y = −log₄(x + 4), labelling where it meets the x- and y-axes and the equation of its asymptote.

  1. 1Vertical asymptote where x + 4 = 0: x = −4
  2. 2y = 0 when x + 4 = 1: (−3, 0)
  3. 3x = 0: y = −log₄ 4 = −1: (0, −1)
  4. 4y → +∞ as x → −4, y decreases as x increases

AnswerDecreasing log curve with vertical asymptote x = −4, falling from +∞ near x = −4, crossing the x-axis at (−3, 0) and the y-axis at (0, −1), then continuing to decrease slowly

Part (b) · 5 marks

Find the value(s) of x satisfying log₍ₓ₊₄₎ 256 − log₄(x + 4) = 0.

  1. 1log₍ₓ₊₄₎ 256 = log₄ 256/log₄(x + 4) = 4/log₄(x + 4)
  2. 24/log₄(x + 4) − log₄(x + 4) = 0
  3. 3(log₄(x + 4))² = 4, so log₄(x + 4) = ±2
  4. 4x + 4 = 16 gives x = 12
  5. 5x + 4 = 1/16 gives x = −63/16

Answerx = 12, x = −63/16

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