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4PM1/02 Jun 2025, Question 7: Exponentials, Logarithms, e and ln

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2 · Calculator · 10 marks

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The question, in short

Complete a table for y = log₃(4 − x) + 3, plot the curve for 0 ≤ x ≤ 3.5, then use a straight line to estimate a root of 3^(2x − 5) − (4 − x)³ = 0.

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

Part (a) · 2 marks

Fill the missing y-values for y = log₃(4 − x) + 3 at the given x-values, rounded to 2 decimal places.

  1. 1x = 0.5: log₃ 3.5 + 3 = 4.14
  2. 2x = 1.5: log₃ 2.5 + 3 = 3.83
  3. 3x = 2.5: log₃ 1.5 + 3 = 3.37

Answerx = 0.5: 4.14; x = 1.5: 3.83; x = 2.5: 3.37

Part (b) · 2 marks

Draw the graph of y = log₃(4 − x) + 3 on the grid for 0 ≤ x ≤ 3.5.

  1. 1Plot the eight points from the table
  2. 2Join them with a smooth curve

AnswerSmooth decreasing curve through the plotted points from (0, 4.26) to (3.5, 2.37)

Part (c) · 6 marks

Draw a suitable straight line and use it to estimate, to 1 decimal place, the root of 3^(2x − 5) − (4 − x)³ = 0 for 0 ≤ x ≤ 3.5.

  1. 13^(2x − 5) = (4 − x)³, so 2x − 5 = 3 log₃(4 − x)
  2. 2(2x − 5)/3 = log₃(4 − x)
  3. 3(2x − 5)/3 + 3 = log₃(4 − x) + 3, i.e. y = (2x + 4)/3
  4. 4Draw y = (2x + 4)/3 on the grid
  5. 5It meets the curve at x ≈ 2.8 (2.7 also accepted)

Answerx = 2.8 (or 2.7)

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