Mathematics 9709/21 — May/June 2018

Walkthrough

The key observation is that $\mathrm{e}^{2x} = (\mathrm{e}^x)^2$, so the equation is quadratic in $\mathrm{e}^x$. To make this clearer, let $u = \mathrm{e}^x$. Then the equation becomes $3u^2 - 82u + 27 = 0$. Factorising gives $(3u - 1)(u - 27) = 0$, so $u = \frac{1}{3}$ or $u = 27$. Because $u = \mathrm{e}^x$ and an exponential is always positive, both values are valid.

Now solve $\mathrm{e}^x = \frac{1}{3}$ and $\mathrm{e}^x = 27$ by taking natural logarithms. Since $\ln(\mathrm{e}^x) = x$, we get $x = \ln\frac{1}{3}$ or $x = \ln 27$. Finally, use the laws of logarithms: $\ln\frac{1}{3} = \ln 3^{-1} = -\ln 3$, and $\ln 27 = \ln 3^3 = 3\ln 3$. These are already in the required form $k\ln 3$.

Key Takeaways

Recognising a hidden quadratic in an exponential equation is essential. Substituting $u = \mathrm{e}^x$ turns the equation into a familiar quadratic, and taking natural logarithms then recovers $x$. The laws $\ln(a^n) = n\ln a$ and $\ln(1/a) = -\ln a$ are needed to write answers in the requested form.

Common Mistakes

• Failing to recognise that $\mathrm{e}^{2x} = (\mathrm{e}^x)^2$ and treating the equation as linear in $x$.
• Stopping after finding $u = \frac{1}{3}$ or $u = 27$, without taking logarithms to find $x$.
• Writing $\ln\frac{1}{3}$ as $\ln 3$ instead of $-\ln 3$.
• Losing one of the two solutions.
• Not showing the substitution or quadratic step, which is needed for the method mark.

Things to Be Careful About

• Both values of $u$ are positive, so there is no extraneous root; however, if a quadratic in $u$ produced a negative value, it would have to be rejected because $\mathrm{e}^x > 0$.
• The mark scheme awards a method mark for attempting the quadratic in $\mathrm{e}^x$ and another method mark for solving $\mathrm{e}^x = c$ where $c > 0$, so show these steps explicitly.
• Give both answers in the form $k\ln 3$; do not leave $\ln\frac{1}{3}$ or $\ln 27$ unsimplified.
• Check that each answer satisfies the original equation.