OCR A Level Mathematics A: Exponential and logarithmic functions: e^x, a^x, ln x and the gradient of e^kx (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 cover exponential and logarithmic functions in OCR A Level Mathematics A (H240), content references 1.06a to 1.06e of the OCR specification, Version 3.1 (August 2026). All of this is Stage 1 content, common to AS and A Level. The marks shown are our own guide to the size of each step. If a topic is unfamiliar, start with the study guide; the revision notes recap each method. Related links: the course hub, the checklist and the Pure Mathematics practice set.

Questions

1. (a) Write 4^(3/2) = 8 in logarithmic form. [1] (b) Evaluate log₂ (1/16). [1] (c) Solve log_x 64 = 3. [1]

2. The curve C has equation y = 3 × 2^x − 5.

(a) Write down the coordinates of the y-intercept. [1] (b) Write down the equation of the asymptote. [1] (c) Describe what happens to y as x → ∞ and as x → −∞. [1] (d) Find the exact x-coordinate of the point where C crosses the x-axis, and give it to 3 significant figures. [2]

3. The populations of two towns, t years from now, are modelled by A = 14000 × 1.02^t and B = 9000 × 1.08^t.

(a) Find A and B when t = 10, to the nearest whole number. [2] (b) Use a table of values to find the first whole number of years after which B is larger than A. [2] (c) Explain why B is larger for all large t. [1]

4. The curve y = e^(−3x) meets the y-axis at P.

(a) Find the gradient of the curve at P and the equation of the tangent at P. [2] (b) Find the equation of the normal at P. [1] (c) The tangent and normal meet the x-axis at Q and R. Find the area of triangle PQR. [2]

5. The mass M mg of a sample after t years is modelled by M = 80e^(−0.04t).

(a) Find dM/dt and its value when t = 0. [2] (b) Show that dM/dt = kM for a constant k, and state k. [1] (c) Find the value of t when the mass is decreasing at 1.6 mg per year, to 3 significant figures. [3]

6. (a) The graph of y = log₃ x is drawn. Find the coordinates of the points on it with y = 2 and y = −1. [2] (b) For the curve y = log₃ (x + 2), state the equation of the asymptote and find the exact coordinates of both axis intercepts. [3]

7. The function f is defined by f(x) = 4e^(x/2) − 3 for all real x.

(a) State the range of f. [1] (b) Find f⁻¹(x). [3] (c) Write down the domain and the range of f⁻¹. [2] (d) Solve f(x) = 9, giving an exact answer. [2]

8. Solve exactly, then give each answer to 3 significant figures: (a) ln(2x − 5) = 3 [2]; (b) e^(3−x) = 10 [2].

9. Without a calculator, write down the value of (a) log₅ 5 [1], (b) e^(ln 7) [1], (c) ln(e^(−2)) [1]. (d) Explain why ln(−3) has no value. [1]

10. The curve C has equation y = 5e^(−x/2) + 1.

(a) Find the coordinates of the y-intercept and the equation of the asymptote. [2] (b) The tangent to C at the y-intercept meets the asymptote at S. Find the coordinates of S. [3] (c) At the point T on C the gradient is −1. Find the x-coordinate of T to 3 significant figures. [3]

Answers

1. (a) log₄ 8 = 3/2 [1] (b) 2⁻⁴ = 1/16, so the value is −4 [1] (c) x³ = 64, so x = 4 [1] Examiner insight: writing the equivalent index statement first, such as x³ = 64, makes it easy to see how the base and the value swap places.

2. (a) x = 0 gives y = 3 − 5, so (0, −2) [1] (b) y = −5 [1] (c) As x → ∞, y → ∞; as x → −∞, y → −5 from above [1] (d) 3 × 2^x = 5, so 2^x = 5/3 [1]; x = log₂ (5/3) = 0.737 [1] Examiner insight: the asymptote is a horizontal line, so it should be given as an equation in y, and not as a value of x.

3. (a) A = 14000 × 1.02¹⁰ = 17 066; B = 9000 × 1.08¹⁰ = 19 430 [2] (b) t = 7: A = 16 082, B = 15 424 [1]; t = 8: A = 16 403, B = 16 658, so 8 years [1] (c) B has the larger growth factor (1.08 > 1.02), so it eventually overtakes and stays ahead [1] Examiner insight: showing two consecutive rows of the table, one on each side of the crossover, is the clearest evidence for a “first whole number” answer.

4. (a) dy/dx = −3e^(−3x); at P(0, 1) the gradient is −3 [1]; tangent y = 1 − 3x [1] (b) Normal gradient 1/3, so y = x/3 + 1 [1] (c) Tangent meets y = 0 at x = 1/3; normal at x = −3 [1]; area = ½ × (1/3 + 3) × 1 = 5/3 [1] Examiner insight: the height is the y-coordinate of P, since QR lies on the x-axis. A quick sketch helps you choose the base and height.

5. (a) dM/dt = 80 × (−0.04)e^(−0.04t) = −3.2e^(−0.04t) [1]; at t = 0 it is −3.2 mg per year [1] (b) −3.2e^(−0.04t) = −0.04 × 80e^(−0.04t), so k = −0.04 [1] (c) −0.04M = −1.6 gives M = 40 [1]; 80e^(−0.04t) = 40, so e^(−0.04t) = 0.5 [1]; t = ln(0.5)/(−0.04) = 17.3 [1] Examiner insight: using the result from (b) turns part (c) into a short calculation. Include units in the rate, and say what the sign means.

6. (a) y = 2: x = 3² = 9, (9, 2) [1]; y = −1: x = 3⁻¹, (1/3, −1) [1] (b) Asymptote x = −2 [1]; x-intercept: x + 2 = 1, (−1, 0) [1]; y-intercept: (0, log₃ 2) [1] Examiner insight: leave log₃ 2 exact when asked for exact coordinates. A decimal can be added, but it does not replace the exact form.

7. (a) e^(x/2) > 0, so f(x) > −3 [1] (b) y + 3 = 4e^(x/2) [1]; x/2 = ln((y + 3)/4) [1]; f⁻¹(x) = 2 ln((x + 3)/4) [1] (c) Domain of f⁻¹ is x > −3; range is all real numbers [2] (d) 4e^(x/2) = 12, so e^(x/2) = 3 [1]; x = 2 ln 3 [1] Examiner insight: the domain of an inverse comes from the range of the original function, so part (a) feeds part (c).

8. (a) 2x − 5 = e³ [1]; x = (e³ + 5)/2 = 12.5 [1] (b) 3 − x = ln 10 [1]; x = 3 − ln 10 = 0.697 [1] Examiner insight: apply the inverse to the whole of each side. In (b), x = ln 10 − 3 would be a sign slip.

9. (a) 1 [1] (b) 7 [1] (c) −2 [1] (d) e^y is positive for every y, so no power of e equals −3 [1] Examiner insight: “without a calculator” invites you to use the inverse relationship, so state it rather than just the number.

10. (a) x = 0 gives y = 6, so (0, 6); asymptote y = 1 [2] (b) dy/dx = −(5/2)e^(−x/2), so the gradient at x = 0 is −5/2 [1]; tangent y = 6 − 5x/2 [1]; y = 1 gives x = 2, so S = (2, 1) [1] (c) −(5/2)e^(−x/2) = −1 [1]; e^(−x/2) = 0.4 [1]; x = −2 ln 0.4 = 1.83 [1] Examiner insight: the constant 1 in the equation of C disappears on differentiating, so it does not affect the gradient. Check the sign in (c).

Where marks are usually lost

• Giving the asymptote of y = a^x + c as y = 0.
• Writing the gradient of e^(kx) without the factor k, or using the wrong sign for a negative k.
• Using the tangent gradient as the normal gradient.
• Leaving an answer in decimal form when it should be exact.
• Rounding too early, so the last digit is wrong.
• Applying ln to part of a side only.
• Forgetting that the domain of an inverse function is the range of the original.
• Taking the logarithm of zero or a negative number.

Next steps

• Review the methods in the revision notes.
• Re-read the study guide for any topic that cost you marks.
• See the OCR A Level Mathematics hub and the checklist.
• Take one of our free 10-minute diagnostics to find your next topic.
• Book a free trial class.

Official syllabus

OCR Level 3 Advanced GCE in Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, published by OCR: section 1.06, Exponentials and Logarithms, references 1.06a to 1.06e.