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WMA12/01 Jun 2019, Question 8: The Laws of Logarithms

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 9 marks

The question, in short

Solve an exponential equation exactly in a specified logarithmic form, then solve a logarithmic equation using logarithm laws.

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

Part (i) · 4 marks

Solve 8^(2x+1) = 6 exactly, giving the answer as a + b log₂3 and finding constants a and b.

  1. 18^(2x+1) = 2^(3(2x+1)) = 2^(6x+3)
  2. 22^(6x+3) = 6, so 6x + 3 = log₂6
  3. 3log₂6 = log₂2 + log₂3 = 1 + log₂3
  4. 46x + 3 = 1 + log₂3, so 6x = −2 + log₂3
  5. 5x = −1/3 + (1/6)log₂3

Answerx = −1/3 + (1/6)log₂3 (a = −1/3, b = 1/6)

Part (ii) · 5 marks

Use logarithm laws to solve log₅(7 − 2y) = 2log₅(y + 1) − 1.

  1. 12log₅(y + 1) = log₅(y + 1)² and 1 = log₅5
  2. 2log₅(7 − 2y) = log₅((y + 1)²/5)
  3. 37 − 2y = (y + 1)²/5, so 35 − 10y = y² + 2y + 1
  4. 4y² + 12y − 34 = 0
  5. 5y = −6 ± √70
  6. 6y = −6 − √70 makes 7 − 2y and y + 1 invalid in the logs, so y = −6 + √70

Answery = −6 + √70

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Also searched as: WMA12 June 2019 · WMA12/01 · wma12 jun19 · IAS Maths Pure Mathematics 2 June 2019 · question 8.
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