A-Level Mathematics May/June 2024 Q2: Solve the equation ln(x-5) = 7-lnx. Give your answer correct to 2 decimal places.

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  The given equation is: $ln(x-5) = 7 - lnx$

  First, rearrange the equation to group the logarithmic terms on one side: $ln(x-5) + lnx = 7$

  Use the product law for logarithms, $ln(a) + ln(b) = ln(ab)$: $ln(x(x-5)) = 7$

  To eliminate the natural logarithm, take the exponential of both sides: $x(x-5) = e^7$

  Expand the brackets and rearrange into a quadratic equation of the form $ax^2 + bx + c = 0$: $x^2 - 5x = e^7$ $x^2 - 5x - e^7 = 0$

  This is a quadratic equation which can be solved using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-e^7)}}{2(1)}$ $x = \frac{5 \pm \sqrt{25 + 4e^7}}{2}$

  Calculating the two possible values for x: $x = \frac{5 + \sqrt{25 + 4e^7}}{2} \quad$ or $\quad x = \frac{5 - \sqrt{25 + 4e^7}}{2}$ $x = 35.710... \quad$ or $\quad x = -30.710...$

  The original equation contains the terms $ln(x)$ and $ln(x-5)$. For these to be defined, we must have $x > 0$ and $x-5 > 0$, which simplifies to $x > 5$.

  Therefore, the solution $x = -30.710...$ is not valid.

  The only valid solution is $x = 35.710...$

  Rounding to 2 decimal places, the final answer is: $x = 35.71$

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  How the marks are awarded

  • M1 — The first mark is for correctly applying a logarithm law. In the model answer, this is achieved by rearranging the equation to ln(x-5) + lnx = 7 and then combining the logarithms into a single term, ln(x(x-5)) = 7.
  • DM1 — This dependent method mark is for correctly eliminating the logarithm after applying the log law. This is shown by converting ln(x(x-5)) = 7 into its exponential form, x(x-5) = e^7.
  • A1 — This accuracy mark is awarded for obtaining a correct quadratic equation in the form ax² + bx + c = 0. The model answer shows this by expanding and rearranging x(x-5) = e^7 to get x² - 5x - e⁷ = 0.
  • A1 — The final accuracy mark is for obtaining the single correct solution x = 35.71. This requires solving the quadratic, rejecting the extraneous negative root (as x must be greater than 5), and rounding the valid root correctly to 2 decimal places.

  Common mistakes

  • Applying an incorrect logarithm law, for example writing ln(x-5) + lnx as ln(x-5+x).
  • Incorrectly eliminating the logarithms from the start, for example changing ln(x-5) = 7 - lnx to x-5 = e^7 - x, without first combining the log terms.
  • Solving the quadratic correctly to find two roots (35.71 and -30.71) but failing to reject the negative solution, which is invalid for the domain of the original logarithmic equation.
  • Forgetting to give the final answer to the required degree of accuracy, or rounding intermediate calculations prematurely.

  Examiner tip: Always aim to combine multiple logarithmic terms into a single logarithm before converting the equation into its exponential form.

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  AI-generated model answer, grounded in the official Cambridge mark scheme and reviewed by the MarkScheme team. Mark your own answer to this question →

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