OCR A Level Mathematics A: Laws of logarithms, exponential equations, linear form and modelling (H240) – 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 follow sections 1.06f to 1.06i of the OCR A Level Mathematics A specification (Version 3.1, August 2026, for first assessment in 2018): laws of logarithms, exponential equations, reduction to linear form and exponential modelling. All four are Stage 1, so they are common to AS and A Level. Pure Mathematics content is assessed in all three written papers.

Read the study guide first, and use the revision notes to recall the methods. The course page is OCR A Level Mathematics and the printable list is the checklist. Give answers to 3 significant figures unless a question says otherwise.

Questions

1. Evaluate log₂ 12 + log₂ 6 − 2 log₂ 3, showing your method. [3]

2. Given that log_a 2 = p and log_a 5 = q, express in terms of p and q:

(a) log_a 50 [2] (b) log_a (8/√5) [2]

3. Solve 2 log₂ x = 3 + log₂ (x + 6). [5]

4. Solve, giving each answer to 3 significant figures:

(a) 7^x = 40 [2] (b) 6^(x+1) = 4^(2x−3) [3]

5.

(a) Solve e^(2x−1) = 15, giving x in exact form. [2] (b) Solve 4^x − 9 × 2^x + 14 = 0, giving each solution exactly or to 3 significant figures. [4]

6. Find x and y given that log₂ x + log₂ y = 5 and y = x + 4. [5]

7. Two variables satisfy y = ax^n. The graph of lg y against lg x is a straight line through (0.2, 0.85) and (0.8, 2.05).

(a) Show that lg y = n lg x + lg a. [2] (b) Find n and a. [4]

8. A quantity y is thought to follow y = kb^x. Measurements are:

x	1	2	3	4	5
y	6.1	9.7	15.6	24.9	39.8

(a) Calculate ln y for each reading, to 2 decimal places. [2] (b) Show that ln y = x ln b + ln k. [1] (c) The points lie close to a straight line. Use the first and last points to estimate b and k. [3] (d) Estimate y when x = 7. [1]

9. The concentration of a drug in the blood is C mg/l, t hours after a dose, where C = C₀e^(−kt). It is 12 at t = 1 and 7.5 at t = 4.

(a) Find k and C₀. [4] (b) Find when the concentration falls to 2 mg/l. [2] (c) Find the half-life. [2]

10. The number of trees in a young woodland is modelled by P = 800e^(0.025t), where t is in years.

(a) Find P when t = 40. [1] (b) Give one limitation of this model for large t, and suggest one refinement. [2]

11. The value £V of an investment after t years is modelled by V = Ae^(rt).

(a) Show that ln V = rt + ln A. [2] (b) Readings of ln V are:

t	2	4	6	8	10
ln V	6.92	7.16	7.40	7.64	7.88

Use the first and last values to find r and A. [4]

(c) Find the time for the investment to double. [2] (d) The investor uses the model for 30 years ahead. Give two reasons for caution. [2]

Answers

1. 2 log₂ 3 = log₂ 9 [1]. Then log₂ (12 × 6 / 9) [1] = log₂ 8 = 3 [1] Examiner insight: Show each law you use. A bare answer of 3 gives the reader no evidence of method, so write the single logarithm before evaluating.

2. (a) 50 = 2 × 5², so log_a 50 = log_a 2 + 2 log_a 5 [1] = p + 2q [1] (b) 8/√5 = 2³ × 5^(−½), so log_a (8/√5) = 3 log_a 2 − ½ log_a 5 [1] = 3p − q/2 [1] Examiner insight: Write the number as a product of powers of 2 and 5 first; trying to apply the laws before factorising is where the working usually goes wrong.

3. 2 log₂ x = log₂ x², and 3 = log₂ 8 [1]. So x² = 8(x + 6) [1]. Then x² − 8x − 48 = 0 [1] and (x − 12)(x + 4) = 0 [1]. x = −4 makes log₂ x undefined, so reject it. x = 12 [1] Examiner insight: A quadratic with two roots invites you to give both. State which root you reject and the reason; leaving it in is a common loss.

4. (a) x ln 7 = ln 40 [1], so x = 3.6889/1.9459 = 1.90 [1] (b) (x + 1) ln 6 = (2x − 3) ln 4 [1]. x (2 ln 4 − ln 6) = 3 ln 4 + ln 6 [1]. x = (3 ln 4 + ln 6)/(2 ln 4 − ln 6) = 6.07 [1] Examiner insight: Keep unrounded values until the last line. Rounded logs inside the fraction can shift the third significant figure.

5. (a) 2x − 1 = ln 15 [1], so x = (1 + ln 15)/2 [1] (b) 4^x = (2^x)², so with t = 2^x: t² − 9t + 14 = 0 [1]. (t − 2)(t − 7) = 0 [1]. 2^x = 2 gives x = 1 [1]. 2^x = 7 gives x = ln 7 / ln 2 = 2.81 [1] Examiner insight: Introduce your substitution in words (“let t = 2^x”). Then convert both values of t back to x, not just one.

6. log₂ (xy) = 5, so xy = 32 [1]. Then x (x + 4) = 32 [1] and x² + 4x − 32 = 0 [1]. (x + 8)(x − 4) = 0 [1]. x = −8 is rejected because log₂ x needs x > 0. x = 4, y = 8 [1] Examiner insight: Test each root in the original logarithmic equations. A pair like (−8, −4) satisfies the algebra but not the log conditions.

7. (a) lg y = lg (ax^n) = lg a + lg x^n [1] = n lg x + lg a [1] (b) Gradient = (2.05 − 0.85)/(0.8 − 0.2) = 2, so n = 2 [1]. Using the point (0.2, 0.85): lg a = 0.85 − 2 × 0.2 [1] = 0.45 [1]. a = 10^0.45 = 2.82 [1] Examiner insight: The intercept is lg a, not a. Marks are usually lost by stopping at 0.45.

8. (a) ln y = 1.81, 2.27, 2.75, 3.21, 3.68 [2] (b) ln (kb^x) = ln k + ln b^x [1] = x ln b + ln k (c) Gradient = (3.68 − 1.81)/4 = 0.4675, so b = e^0.4675 = 1.60 [1]. ln k = 1.81 − 0.4675 = 1.3425 [1], so k = 3.83 [1] (d) y = 3.83 × 1.596^7 ≈ 101 [1] Examiner insight: Keep b and k unrounded when you substitute. Using the rounded 1.60 gives about 103, which shows why you round last.

9. (a) 7.5/12 = e^(−3k), so e^(3k) = 1.6 [1]. k = ln 1.6/3 = 0.157 [1]. C₀ = 12e^k [1] = 14.0 [1] (b) 2 = 14.035e^(−kt) gives e^(kt) = 7.0176 [1], so t = ln 7.0176/0.15667 = 12.4 hours [1] (c) Half-life = ln 2/k [1] = 4.42 hours [1] Examiner insight: Carry the unrounded value of k from part (a) into later parts; re-using 0.157 introduces rounding errors.

10. (a) t = 40: P = 800e = 2170 [1] (b) Limitation: the model grows without bound, but a woodland has limited land and light [1]. Refinement: use a model that levels off at a maximum population [1] Examiner insight: A comment such as “it is not realistic” is too vague. Name the specific feature of the model and tie it to the woodland.

11. (a) ln V = ln (Ae^(rt)) = ln A + ln e^(rt) [1] = rt + ln A [1] (b) Gradient = (7.88 − 6.92)/8 = 0.12 [1], so r = 0.12 [1]. ln A = 6.92 − 2 × 0.12 = 6.68 [1], so A = e^6.68 = 796 [1] (c) Doubling: e^(0.12t) = 2 [1], so t = ln 2/0.12 = 5.78 years [1] (d) The data cover only 2 to 10 years, and 30 years is well outside that range [1]. Interest rates or returns are unlikely to stay at a constant 12% for decades [1] Examiner insight: Give two distinct reasons. Repeating the same point in different words is only one reason.

Where marks are usually lost

• Not stating which root you reject when a logarithm would have a negative input.
• Treating log (x + y) as log x + log y.
• Rounding intermediate logs and then losing accuracy in the final answer.
• Giving the intercept of a log graph as the constant, not its logarithm.
• Mixing bases, for example lg for the gradient and ln for the intercept.
• Writing a bare numerical answer to a “show that” or “evaluate” question with no method.
• Giving vague comments about a model with no link to the context.
• Forgetting units, such as hours or years, in modelling answers.

Next steps

• Read the study guide again for any question you missed.
• Revise from the revision notes.
• Go to the course page and tick off the checklist.
• Try all the free 10-minute diagnostics.
• Finally, book a free trial class.

Official syllabus

OCR, A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, published by OCR. These questions cover section 1.06f to 1.06i.