A-Level Mathematics May/June 2025 Q2: Solve the equation 2 ln (2x+3) - ln(2x + 5) = ln(3x).

1 answer

• ▲0
  accepted ✓

  The given equation is $2 \ln (2x+3) - \ln(2x + 5) = \ln(3x)$.

  First, apply the power law of logarithms to the first term: $2 \ln(2x+3) = \ln((2x+3)^2)$

  Now substitute this back into the equation and apply the quotient law of logarithms to the left-hand side: $\ln((2x+3)^2) - \ln(2x+5) = \ln(3x)$ $\ln\left(\frac{(2x+3)^2}{2x+5}\right) = \ln(3x)$

  Since the logarithms on both sides have the same base, we can equate their arguments: $\frac{(2x+3)^2}{2x+5} = 3x$

  Now, expand the numerator and rearrange to form a quadratic equation: $4x^2 + 12x + 9 = 3x(2x+5)$ $4x^2 + 12x + 9 = 6x^2 + 15x$ $0 = 6x^2 - 4x^2 + 15x - 12x - 9$ $0 = 2x^2 + 3x - 9$

  Solve the quadratic equation by factorisation: $(2x-3)(x+3) = 0$

  This gives two possible solutions for $x$: $2x - 3 = 0 \implies x = \frac{3}{2}$ $x + 3 = 0 \implies x = -3$

  We must check these solutions in the original equation, as the arguments of logarithms must be positive.

  For $x = \frac{3}{2}$:

  • $2x+3 = 2(\frac{3}{2})+3 = 6 > 0$
  • $2x+5 = 2(\frac{3}{2})+5 = 8 > 0$
  • $3x = 3(\frac{3}{2}) = \frac{9}{2} > 0$ All arguments are positive, so $x = \frac{3}{2}$ is a valid solution.

  For $x = -3$:

  • $3x = 3(-3) = -9$. The argument of $\ln(3x)$ is negative. Therefore, $x = -3$ is not a valid solution and must be rejected.

  The only solution is $x = \frac{3}{2}$.

  Final Answer: $x = \frac{3}{2}$

  ---

  How the marks are awarded

  • M1 — Correctly applying a logarithm law. For example, changing 2 ln(2x+3) to ln((2x+3)²), or combining the two terms on the left-hand side to get ln(((2x+3)²)/(2x+5)).
  • A1 — Obtaining a correct equation with no logarithms present, such as (2x+3)² / (2x+5) = 3x.
  • DM1 — Correctly manipulating the logarithm-free equation to form the 3-term quadratic 2x² + 3x - 9 = 0 and proceeding to solve it to find two roots. This mark is dependent on the first M1 mark being awarded.
  • A1 — Stating the single correct final answer x = 3/2 (or 1.5), having clearly rejected the extraneous root x = -3.

  Common mistakes

  • Incorrectly applying logarithm laws, for example writing 2 ln(2x+3) as ln(4x+6) or ln(A) - ln(B) as ln(A-B).
  • Making algebraic errors when expanding brackets or rearranging the equation, for example expanding (2x+3)² to 4x²+9.
  • Failing to check the validity of the solutions. A common error is to state both x = 3/2 and x = -3 as the final answers, without realising that x = -3 makes the argument of a logarithm negative.
  • Correctly forming the quadratic but then making an error in factorisation or when using the quadratic formula.

  Examiner tip: Always check your solutions to logarithmic or radical equations by substituting them back into the original equation, as the algebraic process can introduce extraneous roots.

  ---

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

  @markscheme_answers

Your answer

Sign in to answer this question.