IGCSE Additional Maths · Cambridge 0606 · Edexcel 4PM1

Logarithms and exponentials: notes and questions

Revision for laws of logarithms, change of base, exponential and logarithmic equations, e^x and ln x, graphs and asymptotes, straight-line form, and (4PM1) surds. Both boards. Cambridge 0606 adds e^x, ln x and converting relationships to straight-line form. Edexcel 4PM1 adds simple surds and rationalising the denominator.

Cambridge 0606: section 6, 7.4
Edexcel 4PM1: section 1
14 questions

Practise logarithms and exponentials →

Key points

log_a x = y means a^y = x. So log_a a = 1 and log_a 1 = 0.
Laws: log(xy) = log x + log y; log(x/y) = log x − log y; log(xᵏ) = k log x.
Change of base: log_a x = log_b x ÷ log_b a. In particular log_a b = 1 ÷ log_b a.
To solve aˣ = b, take logs of both sides: x = log b ÷ log a.
Equations like 2^{2x} − 6(2^x) + 8 = 0 become quadratics with u = 2^x. Reject any solution that makes u negative.
y = Axⁿ becomes lg y = n lg x + lg A, a straight line. y = Abˣ becomes lg y = x lg b + lg A.

Watch out: Always check answers in the original equation: a logarithm of zero or a negative number does not exist, so one root of the quadratic often has to go.

Worked example · 2 marks

Solve \(3^x = 20\), giving your answer correct to 3 significant figures.

M1 x = log 20 / log 3 (any base)
A1 2.73

Take logarithms: x ln 3 = ln 20.
x = ln 20 ÷ ln 3 = 2.9957… ÷ 1.0986… = 2.7268…
x = 2.73 to 3 significant figures.

Answer: \(x = 2.73\)

Logarithms and exponentials questions

Original exam-style questions, written for this site and each re-solved independently. Tags show the courses whose content a question tests and whether it can be done without a calculator (Cambridge 0606 Paper 1 is non-calculator).

Question 1 · 3 marks06064PM1no calculator

Without using a calculator, find the value of \(2\lg 5 + \lg 8 - \lg 2\).

Show the mark scheme and worked solution

M1 2 lg 5 = lg 25
M1 combine: lg(25 × 8 ÷ 2)
A1 2

2 lg 5 = lg 5² = lg 25.
lg 25 + lg 8 − lg 2 = lg(25 × 8 ÷ 2) = lg 100.
lg 100 = 2, since 10² = 100.

Answer: \(2\)

Question 2 · 3 marks06064PM1no calculator

Solve \(2^{2x} - 6(2^x) + 8 = 0\).

Show the mark scheme and worked solution

M1 u = 2^x: u² − 6u + 8 = 0
A1 u = 2 or u = 4
A1 x = 1 or x = 2

Let u = 2^x, so 2^{2x} = u²: u² − 6u + 8 = 0.
(u − 2)(u − 4) = 0, so 2^x = 2 or 2^x = 4.
x = 1 or x = 2.

Answer: \(x = 1\) or \(x = 2\)

Question 3 · 4 marks06064PM1no calculator

Given that \(\log_a 2 = p\) and \(\log_a 3 = q\), express in terms of \(p\) and \(q\)

(a) \(\log_a 12\)

(b) \(\log_a 4.5\)

Show the mark scheme and worked solution

M1 (a) 12 = 2² × 3
A1 (a) 2p + q
M1 (b) 4.5 = 3²/2
A1 (b) 2q − p

12 = 2² × 3, so log_a 12 = 2 log_a 2 + log_a 3 = 2p + q.
4.5 = 9/2 = 3²/2, so log_a 4.5 = 2 log_a 3 − log_a 2 = 2q − p.

Answer: (a) \(2p + q\) (b) \(2q - p\)

Question 4 · 3 marks06064PM1no calculator

Solve \(\lg x + \lg(x - 15) = 2\).

Show the mark scheme and worked solution

M1 x(x − 15) = 100
M1 x² − 15x − 100 = 0 solved
A1 x = 20 only

lg(x(x − 15)) = 2, so x(x − 15) = 10² = 100.
x² − 15x − 100 = 0, so (x − 20)(x + 5) = 0.
x = −5 makes lg x undefined, so x = 20.

Answer: \(x = 20\)

9 more logarithms and exponentials questions

Each with a mark scheme and a worked solution. Included with IGCSE plans, free trials and school licences.

Next steps

Cambridge 0606: papers, syllabus map and official past papers
Edexcel 4PM1: papers, syllabus map and official past papers
Going on to A Level? Logarithms and exponentials continues in A Level exponentials and logarithms (A Level Math Revision). See also from Additional Maths to A Level.