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WMA12/01 Jun 2021, Question 7: The Laws of Logarithms

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 10 marks

The question, in short

Solve a logarithmic equation by converting it to a cubic, verify a given root, then solve the original equation algebraically showing all steps.

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

Part (a) · 4 marks

Use log laws to remove logs from 3log₃(2x − 1) = 2 + log₃(14x − 25) and derive the cubic 2x³ − 3x² − 30x + 56 = 0.

  1. 12 = log₃ 9
  2. 2log₃ (2x − 1)³ = log₃ 9(14x − 25)
  3. 3(2x − 1)³ = 9(14x − 25)
  4. 48x³ − 12x² + 6x − 1 = 126x − 225
  5. 58x³ − 12x² − 120x + 224 = 0, divide by 4: 2x³ − 3x² − 30x + 56 = 0

AnswerShown: (2x − 1)³ = 9(14x − 25) ⇒ 8x³ − 12x² − 120x + 224 = 0 ⇒ 2x³ − 3x² − 30x + 56 = 0

Part (b) · 2 marks

Substitute −4 into the cubic and show it satisfies the equation.

  1. 1Substitute x = −4 into 2x³ − 3x² − 30x + 56
  2. 2= −128 − 48 + 120 + 56 = 0, so −4 is a root

AnswerShown: 2(−4)³ − 3(−4)² − 30(−4) + 56 = −128 − 48 + 120 + 56 = 0, so −4 is a root.

Part (c) · 4 marks

Factorise the cubic using the known root and solve the original log equation, checking validity of each solution.

  1. 12x³ − 3x² − 30x + 56 = (x + 4)(2x² − 11x + 14)
  2. 22x² − 11x + 14 = (2x − 7)(x − 2)
  3. 3x = −4, 7/2 or 2
  4. 4x = −4 makes 2x − 1 negative, so reject it
  5. 5x = 2 or x = 7/2

Answerx = 2 or x = 7/2 (x = −4 rejected because the logs are undefined)

Stuck on a step? Read the free lesson: The Laws of Logarithms

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Also searched as: WMA12 June 2021 · WMA12/01 · wma12 jun21 · IAS Maths Pure Mathematics 2 June 2021 · question 7.
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