IGCSE · Practice questions · Solving exponential and logarithmic equations

Take logs, solve, check.

Six original Cambridge-style questions on exponential and logarithmic equations: taking logs, exact and 3 s.f. answers, collecting \(x\) terms, combining logs, change of base, simultaneous equations and growth.

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

The method

Isolate, take logs, solve, check.

Isolate the power: divide by any coefficient so \(a^{x}\) or \(e^{kx}\) stands alone.
Take logs of both sides (\(\ln\) for base \(e\)) and bring the whole power down in a bracket.
Collect the \(x\) terms and factorise out \(x\) when \(x\) is in two powers.
Logs in the equation: change to one base, combine into one log, convert to a power.
Answer and check: exact on Paper 1, 3 s.f. on Paper 2; reject any value that makes a log argument zero or negative.

Recall

[2 marks]

Solve the equation \(6^{x} = 50\), giving your answer to 3 significant figures.

Take \(\lg\): \(x\lg 6 = \lg 50\). ✓
\(x = \dfrac{\lg 50}{\lg 6} = \dfrac{1.6989\ldots}{0.7781\ldots} = 2.1833\ldots\) ✓
\(x = 2.18\). ✓

Calculation

[4 marks]

Solve the equation \(2^{3x - 1} = 5^{x + 1}\), giving your answer (a) in the form \(\dfrac{\ln p}{\ln q}\), where \(p\) and \(q\) are rational, (b) to 3 significant figures.

Take \(\ln\): \((3x - 1)\ln 2 = (x + 1)\ln 5\). ✓
Expand: \(3x\ln 2 - \ln 2 = x\ln 5 + \ln 5\). ✓
Collect: \(x(3\ln 2 - \ln 5) = \ln 5 + \ln 2\), so \(x\ln \dfrac{8}{5} = \ln 10\). ✓
(a) \(x = \dfrac{\ln 10}{\ln 1.6}\). ✓
(b) \(x = \dfrac{2.3025\ldots}{0.4700\ldots} = 4.90\). ✓

Calculation

[4 marks]

Solve the equation \(2\lg x - \lg(x - 2) = \lg 9\).

Power law: \(2\lg x = \lg x^{2}\). Quotient law: \(\lg \dfrac{x^{2}}{x - 2} = \lg 9\). ✓
So \(\dfrac{x^{2}}{x - 2} = 9\), giving \(x^{2} = 9x - 18\) and \(x^{2} - 9x + 18 = 0\). ✓
\((x - 3)(x - 6) = 0\), so \(x = 3\) or \(x = 6\). ✓
Check: both values give \(x > 0\) and \(x - 2 > 0\), so both are valid. ✓

Analysis

[5 marks]

Using the substitution \(u = \log_{3} x\), solve the equation \(\log_{3} x - 4\log_{x} 3 = 3\).

Change of base: \(\log_{x} 3 = \dfrac{\log_{3} 3}{\log_{3} x} = \dfrac{1}{u}\). ✓
The equation becomes \(u - \dfrac{4}{u} = 3\). Multiply by \(u\): \(u^{2} - 3u - 4 = 0\). ✓
\((u - 4)(u + 1) = 0\), so \(\log_{3} x = 4\) or \(\log_{3} x = -1\). A negative log is allowed. ✓
\(x = 3^{4} = 81\) or \(x = 3^{-1} = \dfrac{1}{3}\). ✓
Both are positive and neither is 1, so \(\log_{x} 3\) exists: both are valid. ✓

Analysis

[5 marks]

Solve the simultaneous equations \(3^{x} \times 9^{y} = 243\) and \(\lg x + \lg y = \lg 3\).

\(9^{y} = 3^{2y}\) and \(243 = 3^{5}\), so \(3^{x + 2y} = 3^{5}\) and \(x + 2y = 5\). ✓
\(\lg xy = \lg 3\), so \(xy = 3\), with \(x > 0\) and \(y > 0\). ✓
Substitute \(x = 5 - 2y\): \((5 - 2y)y = 3\), so \(2y^{2} - 5y + 3 = 0\). ✓
\((2y - 3)(y - 1) = 0\), so \(y = 1\) or \(y = \dfrac{3}{2}\). ✓
\(y = 1\) gives \(x = 3\); \(y = \dfrac{3}{2}\) gives \(x = 2\). Both pairs are positive, so both are valid. ✓

Proof

[6 marks]

The population, \(P\), of a town \(t\) years after 1 January 2020 is modelled by \(P = 1200e^{0.035t}\).

(a) State the population on 1 January 2020.

(b) Find the value of \(t\) when \(P = 3000\), giving your answer to 3 significant figures.

(c) Show that the time for the population to double is \(\dfrac{200}{7}\ln 2\) years, whatever the starting time, and evaluate it to 3 significant figures.

(a) At \(t = 0\): \(P = 1200e^{0} = 1200\). ✓
(b) \(1200e^{0.035t} = 3000\), so \(e^{0.035t} = 2.5\) and \(0.035t = \ln 2.5\). ✓
\(t = \dfrac{\ln 2.5}{0.035} = 26.17\ldots\), so \(t = 26.2\). ✓
(c) Doubling from time \(t\) to \(t + T\): \(1200e^{0.035(t + T)} = 2 \times 1200e^{0.035t}\). ✓
Divide by \(1200e^{0.035t}\): \(e^{0.035T} = 2\), so \(T = \dfrac{\ln 2}{0.035} = \dfrac{200}{7}\ln 2\), with no \(t\) left: it does not depend on the start. ✓
\(T = 19.8\) years. ✓

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.