Cambridge International AS & A Level Mathematics 9709: Pure Mathematics 2 Logarithms and Exponentials – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 2.2, Logarithmic and exponential functions, of the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027 (Version 4). Section 2.2 is part of Pure Mathematics 2 and is examined on Paper 2, the AS Level Pure Mathematics route; the same outcomes appear as section 3.2 for Paper 3. A scientific calculator is allowed on every 9709 paper, so all questions are calculator allowed, but unsupported calculator answers earn no marks. Where a question asks for an exact answer, the working must be algebraic. Otherwise give answers to 3 significant figures.

Related: the study guide, the revision notes, the whole-paper Pure Mathematics 2 practice set, the A Level Mathematics hub, the printable 9709 checklist and the free AS diagnostic.

Questions

1. It is given that logₐ 2 = p and logₐ 3 = q. Express each of the following in terms of p and q.

(a) logₐ 12 [1] (b) logₐ (27/16) [2]

2. Solve the equation 7^(2x + 1) = 30, giving x correct to 3 significant figures. [3]

3. Solve the equation ln(x + 4) − ln x = 2 ln 3, giving your answer in exact form. [3]

4.

(a) Sketch, on the same diagram, the graphs of y = e^(0.5x) and y = e^(−2x). State the coordinates of any point where either graph meets an axis, and the equation of the asymptote. [3] (b) Find the x-coordinate of the point where the curves y = e^(0.5x) and y = 4e^(−2x) meet, giving your answer correct to 3 significant figures. [2]

5. Solve the equation 3^(2x) − 4(3^(x + 1)) + 27 = 0. [5]

6. Find the smallest integer n such that 1 − 0.85ⁿ > 0.99. [4]

7.

(a) Show that the solution of the equation 2^(3x − 1) = 5^(x + 1) can be written as x = (ln 10)/(ln 1.6). [3] (b) Hence find x correct to 3 significant figures. [1]

8. The variables x and y satisfy the equation y = kxⁿ, where k and n are constants.

(a) Show that the graph of ln y against ln x is a straight line, and state its gradient and its intercept on the ln y axis in terms of k and n. [2] (b) The line passes through the points (0.693, 2.203) and (1.792, 3.851). Find the values of n and k, giving each correct to 3 significant figures. [3] (c) Use your values to estimate y when x = 4. [1]

9. The variables x and y satisfy the equation y = k(aˣ), where k and a are constants. The graph of ln y against x is a straight line passing through the points (2, 4.1) and (7, 2.6).

(a) Find the values of a and k, giving each correct to 3 significant figures. [4] (b) Find the value of x for which y = 20. [2]

10. Solve the equation 2 ln(x + 1) − ln(x + 3) = ln 2, giving your answer in exact form. [6]

11.

(a) Show that, using the substitution u = eˣ, the equation 3eˣ + 2e^(−x) = 7 becomes 3u² − 7u + 2 = 0. [2] (b) Hence solve 3eˣ + 2e^(−x) = 7, giving your answers in exact form. [3] (c) Solve the inequality 3eˣ + 2e^(−x) < 7. [2] (d) Hence solve 3e^(2y) + 2e^(−2y) = 7, giving your answers in exact form. [2]

12. The mass, m grams, of a radioactive sample t days after it is prepared is modelled by m = m₀(aᵗ), where m₀ and a are constants. The mass is 57.8 grams when t = 2 and 35.5 grams when t = 5.

(a) Show that ln m is a linear function of t. [1] (b) Find the values of a and m₀, giving each correct to 3 significant figures. [4] (c) Using a = 0.85 and m₀ = 80, find the number of complete days after which the mass first falls below 1 gram. [3]

Answers

1. (a) logₐ 12 = logₐ(2² × 3) = 2p + q [1] (b) logₐ(27/16) = logₐ 27 − logₐ 16 = 3 logₐ 3 − 4 logₐ 2 [1] = 3q − 4p [1] Examiner insight: 27/16 must be split as 3³ ÷ 2⁴; writing logₐ 27 ÷ logₐ 16 is a wrong law and loses both marks in (b).

2. (2x + 1) ln 7 = ln 30 [1]. 2x + 1 = 3.4012/1.9459 = 1.7479 [1]. x = 0.7479/2 = 0.374 [1] Examiner insight: the method mark needs the whole bracket (2x + 1) multiplying ln 7; writing 2x + 1 ln 7 loses it, even if the final number is right.

3. ln((x + 4)/x) = ln 9, using 2 ln 3 = ln 9 [1]. (x + 4)/x = 9, so x + 4 = 9x [1]. x = 1/2 [1] Examiner insight: “exact form” means 1/2 (or 0.5); an answer found by trial on a calculator with no log working earns nothing.

4. (a) y = e^(0.5x): increasing curve through (0, 1) [1]. y = e^(−2x): decreasing curve through (0, 1), steeper near the y-axis [1]. Both approach the asymptote y = 0 without touching it; (0, 1) labelled and y = 0 stated [1] (b) e^(0.5x) = 4e^(−2x) ⇒ e^(2.5x) = 4 ⇒ 2.5x = ln 4 [1]. x = 1.3863/2.5 = 0.555 [1] Examiner insight: a curve that crosses or clearly touches the x-axis loses the asymptote mark, however good the rest of the sketch is.

5. 3^(x + 1) = 3 × 3ˣ [1]. Let u = 3ˣ: u² − 12u + 27 = 0 [1]. (u − 3)(u − 9) = 0, so u = 3 or u = 9 [1]. 3ˣ = 3 gives x = 1 [1]; 3ˣ = 9 gives x = 2 [1] Examiner insight: treating 3^(x + 1) as 3ˣ + 1 or 3ˣ⁺¹ = 3ˣ gives the wrong quadratic, and no later marks can be earned from it.

6. 0.85ⁿ < 0.01 [1]. n ln 0.85 < ln 0.01 [1]. ln 0.85 < 0, so the inequality reverses: n > (ln 0.01)/(ln 0.85) = 28.34 [1]. Smallest integer n = 29 [1] Examiner insight: the third mark is for reversing the sign with a reason; “n < 28.3” followed by “n = 29” is inconsistent and loses it.

7. (a) (3x − 1) ln 2 = (x + 1) ln 5 [1]. 3x ln 2 − x ln 5 = ln 5 + ln 2, so x(3 ln 2 − ln 5) = ln 5 + ln 2 [1]. ln 5 + ln 2 = ln 10 and 3 ln 2 − ln 5 = ln(8/5) = ln 1.6, so x = (ln 10)/(ln 1.6) [1] (b) x = 2.3026/0.4700 = 4.90 [1] Examiner insight: this is a “show that” with the answer given, so the log laws turning each side into ln 10 and ln 1.6 must be written; jumping to the result loses the final mark.

8. (a) ln y = ln(kxⁿ) = ln k + n ln x [1]. This is of the form Y = c + mX with Y = ln y and X = ln x, so it is a straight line with gradient n and intercept ln k [1] (b) n = (3.851 − 2.203)/(1.792 − 0.693) = 1.648/1.099 [1] = 1.50 [1]. ln k = 2.203 − 1.4995 × 0.693 = 1.164, so k = e^1.164 = 3.20 [1] (c) y = 3.20 × 4^1.50 = 25.6 [1] Examiner insight: stating the intercept is k rather than ln k in (a) loses the mark there and usually the k mark in (b) as well.

9. (a) ln y = ln k + x ln a [1]. Gradient = (2.6 − 4.1)/(7 − 2) = −0.3 = ln a, so a = e^(−0.3) = 0.741 [1]. 4.1 = ln k + 2(−0.3), so ln k = 4.7 [1]. k = e^4.7 = 110 [1] (b) ln 20 = 4.7 − 0.3x [1]. x = (4.7 − 2.9957)/0.3 = 5.68 [1] Examiner insight: use the exact line ln y = 4.7 − 0.3x in (b); working from the rounded a and k is accepted only if the final answer still rounds to 5.68.

10. 2 ln(x + 1) = ln(x + 1)² [1]. ln((x + 1)²/(x + 3)) = ln 2 [1]. (x + 1)² = 2(x + 3) [1]. x² + 2x + 1 = 2x + 6, so x² = 5 [1]. x = √5 or x = −√5 [1]. x = −√5 makes x + 1 negative, so ln(x + 1) is undefined; x = √5 [1] Examiner insight: the last mark needs the rejection of −√5 with a reason; leaving “x = ±√5” loses it.

11. (a) Multiply by eˣ: 3e^(2x) + 2 = 7eˣ [1]. With u = eˣ, 3u² + 2 = 7u, so 3u² − 7u + 2 = 0 [1] (b) (3u − 1)(u − 2) = 0, so u = 1/3 or u = 2 [1]. eˣ = 1/3 gives x = −ln 3 [1]; eˣ = 2 gives x = ln 2 [1] (c) eˣ > 0, so multiplying by eˣ keeps the sign: 3u² − 7u + 2 < 0, so 1/3 < eˣ < 2 [1]. −ln 3 < x < ln 2 [1] (d) Here 2y plays the part of x, so 2y = −ln 3 or 2y = ln 2 [1]. y = −½ ln 3 or y = ½ ln 2 [1] Examiner insight: ln(1/3) is accepted for −ln 3 in (b), but in (c) giving x < ln 2 alone, or the region outside the roots, earns only the first mark.

12. (a) ln m = ln(m₀aᵗ) = ln m₀ + t ln a, which is linear in t with gradient ln a and intercept ln m₀ [1] (b) ln 57.8 = ln m₀ + 2 ln a and ln 35.5 = ln m₀ + 5 ln a [1]. Subtracting: 3 ln a = ln(35.5/57.8) = −0.4875, so ln a = −0.1625 [1]. a = 0.850 [1]. ln m₀ = 4.0570 + 0.3250 = 4.382, so m₀ = 80.0 [1] (c) 80(0.85ᵗ) < 1, so t ln 0.85 < ln(1/80) [1]. ln 0.85 < 0, so t > (ln 0.0125)/(ln 0.85) = 26.96 [1]. The mass first falls below 1 gram after 27 days [1] Examiner insight: in (c), an answer of 26 days (rounding down) loses the final mark: after 26 days the mass is still 1.17 grams.

Where marks are usually lost

• Brackets dropped when the power law is used, so (2x + 1) ln 7 becomes 2x + ln 7.
• Logs removed term by term instead of first combining into a single log on each side.
• 3^(x + 1) not rewritten as 3 × 3ˣ, so the hidden quadratic is never found.
• Inequality signs not reversed after dividing by the log of a number between 0 and 1.
• Roots that make a log undefined, or eˣ negative, kept in the final answer, or rejected with no reason.
• The intercept of a ln y graph given as k instead of ln k.
• Premature rounding of a gradient, so the constant found from e^(gradient) is wrong at 3 s.f.
• “Show that” answers that skip the log-law step turning the result into the given form.
• Sketches of y = e^(kx) that cross the x-axis or omit (0, 1).

Next steps

• Logarithms and exponentials revision notes
• Logarithms and exponentials study guide
• Pure Mathematics 2 guide
• A Level Mathematics hub
• Printable 9709 checklist
• AS diagnostic, the 9709 self-check bank, or all free 10-minute diagnostics
• Book a free trial class

Official syllabus

Cambridge International AS & A Level Mathematics 9709 syllabus, for exams in 2026 and 2027 (Version 4), Cambridge University Press & Assessment. Topic 2, Pure Mathematics 2 (for Paper 2): section 2.2 Logarithmic and exponential functions.