AS & A Level · Practice questions · The exponential function and natural logarithms

Growth, decay, undo.

Papers 2 and 3. Six original Cambridge-style questions on \(e^x\) and \(\ln x\), finishing with Newton-style cooling in context.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Isolate, then apply the inverse.

Inverse pair: \(\ln(e^x) = x\) and \(e^{\ln x} = x\) (\(x > 0\)).
Isolate first: get \(e^{(\dots)}\) or \(\ln(\dots)\) alone on one side.
Quadratic in \(e^x\): let \(u = e^x\); reject \(u \le 0\).
Sketch: \(y\)-intercept, asymptote \(y = c\) for \(Ae^{kx} + c\), and any \(x\)-intercept.
Exact or 3 s.f.: give \(\ln\) forms when asked for exact answers.

Recall

[4 marks]

Simplify exactly: (a) \(e^{3\ln 2}\) (b) \(\ln \dfrac{1}{e^2}\) (c) \(e^{\ln 6 - \ln 3}\) (d) \(\ln\left(e^{2x} \times e^3\right)\).

(a) \(e^{\ln 8} = 8\) ✓
(b) \(\ln e^{-2} = -2\) ✓
(c) \(e^{\ln 2} = 2\) ✓
(d) \(\ln e^{2x + 3} = 2x + 3\) ✓

Calculation

[4 marks]

Solve, giving answers exactly and to 3 significant figures: (a) \(e^{3x} = 20\) (b) \(\ln(2x - 1) = 3\).

(a) \(3x = \ln 20\), so \(x = \dfrac{\ln 20}{3} = 0.999\) (3 s.f.). ✓
(b) \(2x - 1 = e^3\), so \(x = \dfrac{e^3 + 1}{2} = 10.5\) (3 s.f.). ✓

Calculation

[3 marks]

Sketch the graph of \(y = 2e^{-x} - 1\), stating the exact coordinates of the points where it meets the axes and the equation of the asymptote.

\(x = 0\): \(y = 2 - 1 = 1\), so \((0, 1)\). ✓
\(y = 0\): \(e^{-x} = \dfrac{1}{2}\), so \(x = \ln 2\): \((\ln 2, 0)\). ✓
As \(x \to \infty\), \(y \to -1\): asymptote \(y = -1\). Decreasing curve, approaching \(y = -1\) from above. ✓

Analysis

[4 marks]

Solve the equation \(3e^{2x} + e^x - 2 = 0\), giving your answer in exact form.

Let \(u = e^x\): \(3u^2 + u - 2 = 0\). ✓
\((3u - 2)(u + 1) = 0\), so \(u = \dfrac{2}{3}\) or \(u = -1\). ✓
\(e^x = -1\) is impossible, so \(e^x = \dfrac{2}{3}\). ✓
\(x = \ln \dfrac{2}{3}\) \((\approx -0.405)\). ✓

Analysis

[4 marks]

Solve the equation \(2\ln(x + 1) - \ln x = \ln 6\), giving your answers in exact form.

\(\ln \dfrac{(x + 1)^2}{x} = \ln 6\), so \((x + 1)^2 = 6x\). ✓
\(x^2 + 2x + 1 = 6x\), so \(x^2 - 4x + 1 = 0\). ✓
\(x = \dfrac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}\). ✓
Both are positive (\(2 - \sqrt{3} \approx 0.27\)), so both logs are defined: \(x = 2 + \sqrt{3}\) or \(x = 2 - \sqrt{3}\). ✓

Analysis

[6 marks]

A cup of tea cools so that its temperature \(T\,\degree\text{C}\) after \(t\) minutes is \(T = 20 + 70e^{-0.05t}\).

(a) Find the initial temperature.

(b) Find the temperature after 10 minutes.

(d) State the temperature that the tea approaches in the long term.

(a) \(t = 0\): \(T = 20 + 70 = 90\,\degree\text{C}\). ✓
(b) \(T = 20 + 70e^{-0.5} = 62.5\,\degree\text{C}\) (3 s.f.). ✓
(c) \(70e^{-0.05t} = 20\), so \(e^{-0.05t} = \dfrac{2}{7}\) and \(-0.05t = \ln \dfrac{2}{7}\). ✓
\(t = 20\ln \dfrac{7}{2} = 25.1\) minutes (3 s.f.). ✓
(d) As \(t \to \infty\), \(e^{-0.05t} \to 0\), so \(T \to 20\,\degree\text{C}\) (room temperature). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.