e^x and ln x

$e \approx 2.718$ is the natural base, and $\ln x$ means $\log_e x$. Everything 0606 asks about the pair follows from one fact: they are inverse functions, each undoes the other.

$e^{\ln x} = x$ and $\ln(e^x) = x$

Solving: apply the other one

$e^{2x-1} = 5$ → take $\ln$ of both sides: $2x - 1 = \ln 5$ → $x = \dfrac{1 + \ln 5}{2}$ $\ln(3x + 2) = 2$ → exponentiate: $3x + 2 = e^2$ → $x = \dfrac{e^2 - 2}{3}$

Write the “apply $\ln$/$e$ to both sides” step explicitly, it’s the method mark. On Paper 1, answers stay in exact form: $\dfrac{1 + \ln 5}{2}$ is finished; $1.30$ is an approximation the question didn’t ask for and can cost the accuracy mark. On Paper 2, evaluate only if the question requests decimals.

All log laws hold for $\ln$ (it’s just base $e$): $\ln a + \ln b = \ln ab$, $n \ln a = \ln a^n$. The disguised-quadratic pattern appears here too: $e^{2x} - 5e^x + 6 = 0$ → $u = e^x$ → $u^2 - 5u + 6 = 0$ → $u = 2, 3$ → $x = \ln 2, \ln 3$, both valid since both $u$-values are positive (an $e^x$-substitution can only accept positive $u$, and saying so when rejecting is a mark).

The graphs

• $y = e^x$: through $(0, 1)$, always positive, $x$-axis asymptote as $x \to -\infty$
• $y = \ln x$: through $(1, 0)$, defined only for $x > 0$, $y$-axis asymptote
• Each is the other reflected in $y = x$

Sketch marks: the labelled intercept and the asymptote drawn/stated. Transformations move them, $y = e^x - 4$ has asymptote $y = -4$ and range $f(x) > -4$ (range from asymptote).

Where e shows up across the paper

The pair is syllabus glue: calculus ($e^x$ differentiates to itself; $\ln x$ to $\dfrac{1}{x}$; $\int \frac{1}{x}\, dx$ involves $\ln$), exponential growth/decay models ($P = P_0 e^{kt}$, solve for $t$ with $\ln$), and linear-form reductions using $\ln$ instead of $\log$. A wobble here taxes three other topics.

Common mistakes

• $\ln$ treated as base-10 log (or the calculator’s log button used for $\ln$)
• Exact answers decimalised on Paper 1
• $\ln$ applied to one side only
• Disguised quadratics solved to $u$ and abandoned before returning to $x$
• Negative $u$-values “solved” as $x = \ln(\text{negative})$, undefined, must be rejected in writing

Full topic context: Logs & Exponentials notes.