Mathematics 9709 · syllabus 2.2, 3.2

Cambridge A Level Mathematics 9709 — Logarithmic and exponential functions topical past papers

Logarithms and their laws, the functions eˣ and ln x, equations with the unknown in a power, and reducing curves to straight-line form.

Questions: 100
Years: 2009–2026
With mark scheme: 100%
Questions 2022–2026: 32

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What Logarithmic and exponential functions covers

A logarithm answers the question 'what power?': logₐ b = c means aᶜ = b. Papers 2 and 3 (sections 2.2 and 3.2, which have the same content) use the laws of logarithms, though not the change-of-base formula, to simplify expressions and to solve equations and inequalities in which the unknown is a power, such as 5^(2x) = 8 or 4 × 3ˣ ≥ 50. The natural exponential eˣ and the natural logarithm ln x are inverse functions; you need their graphs, including y = e^(kx) for positive and negative k.

The other main skill is turning a relationship into a straight line. Taking logarithms of y = kxⁿ gives ln y = ln k + n ln x, a straight line when ln y is plotted against ln x; taking logarithms of y = kaˣ gives ln y = ln k + x ln a, a straight line when ln y is plotted against x. From a given line's gradient and intercept you then find the unknown constants.

Exponentials and logarithms are needed throughout the rest of Papers 2 and 3: differentiating and integrating eˣ and ln x, and solving differential equations whose solutions are exponential. Equations that are quadratic in eˣ or 2ˣ link back to the quadratics of Paper 1, and eˣ and ln x are a particular case of a one-one function and its inverse from Paper 1, so each graph is the reflection of the other in the line y = x.

In the syllabus

The headings this topic has in the Cambridge International 2026–2027 syllabus, as the board prints them.

2.2 Logarithmic and exponential functions
3.2 Logarithmic and exponential functions

Read the syllabus on the board’s website

What you should be able to do

Convert between logarithmic and index form
Use the laws of logarithms to combine and simplify expressions
Solve equations and inequalities with the unknown in an index, using logarithms
Sketch y = eˣ, y = e^(kx) and y = ln x, and use their inverse relationship
Transform y = kxⁿ or y = kaˣ to linear form and find the constants from a graph
Solve equations that are quadratic in eˣ or in aˣ

Key ideas

logₐ b = c ⇔ aᶜ = b; ln x means logₑ x
log(xy) = log x + log y; log(x/y) = log x − log y; log(xⁿ) = n log x
e^(ln x) = x for x > 0, and ln(eˣ) = x for all x
y = kxⁿ ⇔ ln y = ln k + n ln x: a straight line in ln x and ln y with gradient n
y = kaˣ ⇔ ln y = ln k + x ln a: a straight line in x and ln y with gradient ln a

How Logarithmic and exponential functions has been examined

From 2022 to 2026, Logarithmic and exponential functions was on Paper 3 · Pure Mathematics 3 in 10 of 11 exam series, worth about 6% of that paper’s marks.

How Logarithmic and exponential functions has been examined, by paper
Paper	Questions	Share of marks	Series
Paper 3 · Pure Mathematics 3	32	6%	10 of 11

Momentum’s own count, over the questions in its bank for these years: a series is one exam session, every zone included, and a question filed under two chapters counts in both. Median marks per question: 4.

Filed in Momentum under: Logarithmic & Exponential Functions.

A worked example

Solve the equation 5^(2x) − 4 × 5ˣ − 12 = 0, giving x correct to 3 significant figures.

Let u = 5ˣ, so 5^(2x) = u² and the equation becomes u² − 4u − 12 = 0.
Factorise: (u − 6)(u + 2) = 0, so u = 6 or u = −2.
5ˣ is always positive, so u = −2 gives no solution.
5ˣ = 6, so x ln 5 = ln 6 and x = ln 6 / ln 5 = 1.7918 / 1.6094 = 1.1133.

Answer: x = 1.11 (3 s.f.)

Written by Momentum for this page, with its own numbers; it is not taken from a past paper.

Mistakes that cost marks

Writing ln(a + b) = ln a + ln b: the law applies to a product, ln(ab), not a sum.
Dividing an inequality by a negative logarithm, such as ln 0.5, without reversing it: logarithms of numbers between 0 and 1 are negative.
Keeping an impossible root from a quadratic in eˣ, such as eˣ = −2: eˣ is always positive, so it gives no solution and must be rejected.
Reading the intercept of a graph of ln y against x as k rather than ln k: the intercept has to be exponentiated to give k.

Exam technique

Take logarithms of both sides to bring the unknown down, collect the x terms, then divide; give x to 3 significant figures unless an exact form is asked for.
In linear-form questions, write down which variable is on each axis, then match the gradient and intercept to the equation term by term.
Use a substitution such as u = eˣ to show a quadratic structure, and reject any negative u before taking logarithms.

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