A-Level Mathematics May/June 2024 Q2: Use logarithms to solve the equation 6^(2x−1) = 5e^(3x+2). Give your answer correct to…

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  The given equation is: $6^{2x-1} = 5e^{3x+2}$

  To solve for $x$, we take the natural logarithm (ln) of both sides: $\ln(6^{2x-1}) = \ln(5e^{3x+2})$

  Using the power rule for logarithms ($ln(a^b) = bln(a)$) on the left side, and the product rule ($ln(ab) = ln(a) + ln(b)$) on the right side: $(2x-1)\ln(6) = \ln(5) + \ln(e^{3x+2})$

  Since $\ln(e^y) = y$, the equation simplifies to: $(2x-1)\ln(6) = \ln(5) + 3x + 2$

  Now, we expand the bracket and rearrange to solve the linear equation for $x$. $2x\ln(6) - \ln(6) = \ln(5) + 3x + 2$

  Group terms containing $x$ on one side and constant terms on the other: $2x\ln(6) - 3x = \ln(5) + 2 + \ln(6)$

  Factor out $x$ on the left side: $x(2\ln(6) - 3) = \ln(5) + 2 + \ln(6)$

  Isolate $x$ by dividing: $x = \frac{\ln(5) + 2 + \ln(6)}{2\ln(6) - 3}$

  Now, we calculate the value using a calculator: $x = \frac{1.6094379... + 2 + 1.791759...}{2(1.791759...) - 3}$ $x = \frac{5.401197...}{3.583518... - 3}$ $x = \frac{5.401197...}{0.583518...}$ $x = 9.25640...$

  Rounding the answer to 4 significant figures: $x = 9.256$

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  How the marks are awarded

  • M1 — The first method mark is awarded for correctly applying the power rule of logarithms to the left-hand side of the equation, resulting in the term $(2x-1)\ln(6)$.
  • *M1 — The second method mark is awarded for correctly applying the product rule and the inverse property of logarithms to the right-hand side, resulting in the expression $\ln(5) + 3x + 2$.
  • DM1 — This dependent method mark is awarded for a correct attempt to solve the linear equation for $x$. This involves expanding brackets, collecting all $x$ terms on one side and all constant terms on the other, and factorising out $x$.
  • A1 — The final accuracy mark is awarded for obtaining the correct answer $9.256$, correctly rounded to 4 significant figures from a correct working.

  Common mistakes

  • Incorrectly applying the logarithm product rule to the right-hand side, for example writing $\ln(5e^{3x+2})$ as $\ln(5) \times \ln(e^{3x+2})$.
  • Making an algebraic error when expanding the bracket, such as writing $(2x-1)\ln(6)$ as $2x\ln(6) - 1$ instead of $2x\ln(6) - \ln(6)$.
  • Errors in rearranging the linear equation, for example moving $3x$ to the left side but forgetting to change its sign.
  • Rounding intermediate values (like $\ln(6)$ and $\ln(5)$) too early, which can lead to an inaccurate final answer that is not awarded the A1 mark.

  Examiner tip: When an equation has the unknown variable in the exponent of different bases, the key technique is to take logarithms of both sides to bring the exponents down and form a solvable linear equation.

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