AQA A-Level Mathematics: F: Exponentials and logarithms (7357) – 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 F: Exponentials and logarithms (F1 to F7) of the AQA A-level Mathematics (7357) specification, version 1.3 (31 January 2018), for A-level exams from June 2018 onwards. Section F is Paper 1 content, and Papers 2 and 3 can also assess it. A calculator is required in every 7357 paper, so all questions assume one, but questions marked (exact) need answers in terms of ln, e or logs. Give other answers to 3 significant figures.

Learn the methods first in the study guide and review with the revision notes. Course hub: AQA A-Level Mathematics · printable checklist · free diagnostics

Questions

1. Write 3 log_a 2 + log_a 5 − ½ log_a 16 as a single logarithm in its simplest form. [3]

2. Solve 7ˣ = 30. [2]

3. A curve has equation y = 2 + ln(x − 1), for x > 1.

(a) State the equation of the asymptote to the curve. [1] (b) Find the exact x-coordinate of the point where the curve crosses the x-axis. (exact) [2] (c) Find the exact x-coordinate of the point on the curve where y = 4. (exact) [1]

4. Solve log₃(2x + 1) − log₃(x − 1) = 2. (exact) [4]

5. Solve 3^(2x + 1) − 10 × 3ˣ + 3 = 0. [5]

6. Solve 5^(x + 1) = 3^(2x). Give your answer in the form ln p / ln q, where p and q are rational, and then as a decimal. [4]

7. The number of fish in a lake is modelled by P = 500e^(0.04t), where t is the time in years after the fish were introduced.

(a) Show that dP/dt = 0.04P. [2] (b) Find the rate of increase of the population when P = 1500. [1] (c) Find the time taken for the population to reach 1500. [2]

8. A scientist believes that two quantities are related by y = axⁿ. She records:

x	2	5	10	20
y	5.0	10.6	18.9	33.7

(a) Show that the model gives a straight line when ln y is plotted against ln x, and state what the gradient and intercept represent. [2] (b) Calculate ln x and ln y for x = 2 and x = 20, to 3 decimal places. [2] (c) Assume the points lie close to a straight line. Use your values from (b) to estimate n and a. [3] (d) Use your model to estimate y when x = 15. [1]

9. The number of bacteria N in a sample, t hours after the start of an experiment, is recorded. A graph of log₁₀ N against t is a straight line with intercept 2.3 and gradient 0.12.

(a) Show that N = kbᵗ, and find k and b. [3] (b) Interpret the value of b in context. [1] (c) Find the time at which the model predicts 10 000 bacteria. [2]

10. After a dose, the concentration of a drug in a patient’s blood is modelled by C = 12e^(−kt) mg/L, where t is in hours. The concentration halves every 4 hours.

(a) Show that k = ¼ ln 2. [2] (b) Find the time after the dose at which the concentration falls to 1.5 mg/L. [3] (c) Find the rate at which the concentration is decreasing when t = 2. [2]

11. The population of a town was 24 000 in 2010 and 31 000 in 2020. It is modelled by P = Ae^(kt), where t is the number of years after 2010.

(a) Find A and k. [3] (b) Use the model to predict the population in 2040. [2] (c) Find the year in which the model first predicts a population above 50 000. [2] (d) Give one limitation of this model for long-term prediction, and suggest one refinement. [2]

Answers

1. 3 log_a 2 = log_a 8 [1]; ½ log_a 16 = log_a 4 [1]; log_a (8 × 5 / 4) = log_a 10 [1] [3] Examiner insight: the power law must be applied to each coefficient before combining; a final answer left as log_a (40/4) usually loses the last mark for “simplest form”.

2. x ln 7 = ln 30, so x = ln 30 / ln 7 [1] = 1.75 (3 s.f.) [1] [2] Examiner insight: a bare 1.75 from a calculator’s solve function risks the method mark; show the logarithm step.

3. (a) x = 1 [1] (b) 2 + ln(x − 1) = 0, so ln(x − 1) = −2 [1]; x = 1 + e⁻² [1] (c) ln(x − 1) = 2, so x = 1 + e² [1] Examiner insight: “exact” means 1 + e⁻², not 1.135; a decimal here gets the method mark but not the accuracy mark.

4. log₃ [(2x + 1)/(x − 1)] = 2 [1]; (2x + 1)/(x − 1) = 3² = 9 [1]; 2x + 1 = 9x − 9, so x = 10/7 [1]; check: 10/7 > 1, so both log arguments are positive and the root is valid [1] Examiner insight: removing the logs as if 2 = log₃ 2 (writing (2x + 1)/(x − 1) = 2) is a common slip and loses every mark after the first.

5. 3^(2x + 1) = 3 × (3ˣ)², so with u = 3ˣ [1]: 3u² − 10u + 3 = 0 [1]; (3u − 1)(u − 3) = 0, u = 1/3 or u = 3 [1]; 3ˣ = 1/3 gives x = −1 [1]; 3ˣ = 3 gives x = 1 [1] Examiner insight: writing 3^(2x + 1) as (3ˣ)² + 3 or 9ˣ alone is the usual error; the factor 3 from 3¹ must stay.

6. (x + 1) ln 5 = 2x ln 3 [1]; x ln 5 − 2x ln 3 = −ln 5, so x(2 ln 3 − ln 5) = ln 5 [1]; x = ln 5 / (ln 9 − ln 5) = ln 5 / ln(9/5) [1] = 2.74 (3 s.f.) [1] Examiner insight: brackets matter: writing x + 1 ln 5 instead of (x + 1) ln 5 changes the equation, so the working that follows can lose its method marks.

7. (a) dP/dt = 500 × 0.04e^(0.04t) [1] = 0.04 × 500e^(0.04t) = 0.04P [1] (b) dP/dt = 0.04 × 1500 = 60 fish per year [1] (c) e^(0.04t) = 3 [1]; t = ln 3 / 0.04 = 27.5 years (3 s.f.) [1] Examiner insight: in a “show that” the final line must read dP/dt = 0.04P explicitly; stopping at 20e^(0.04t) does not earn the second mark.

8. (a) ln y = ln a + ln xⁿ = ln a + n ln x [1]; this is Y = c + mX with gradient n and intercept ln a [1] (b) x = 2: ln x = 0.693, ln y = 1.609 [1]; x = 20: ln x = 2.996, ln y = 3.517 [1] (c) n = (3.517 − 1.609)/(2.996 − 0.693) [1] = 0.828 [1]; ln a = 1.609 − 0.828 × 0.693 = 1.035, a = e^1.035 = 2.82 [1] (d) y = 2.82 × 15^0.828 = 26.5 [1] Examiner insight: the intercept is ln a, so quoting a = 1.035 loses the accuracy mark; follow-through usually applies to (d) from your own a and n.

9. (a) log₁₀ N = 2.3 + 0.12t, so N = 10^2.3 × (10^0.12)ᵗ [1]; k = 10^2.3 = 200 (3 s.f.) [1]; b = 10^0.12 = 1.32 (3 s.f.) [1] (b) The number of bacteria is multiplied by about 1.32 each hour, an increase of about 32% per hour [1] (c) log₁₀ 10 000 = 4, so 4 = 2.3 + 0.12t [1]; t = 14.2 hours (3 s.f.) [1] Examiner insight: interpreting b needs the context and the time unit (“per hour”); “b is the growth rate” alone is too vague for the mark.

10. (a) 12e^(−4k) = 6 [1]; e^(−4k) = ½, so −4k = ln ½ = −ln 2, k = ¼ ln 2 [1] (b) 12e^(−kt) = 1.5, so e^(−kt) = 1/8 [1]; t = ln 8 / k [1] = 3 ln 2 / (¼ ln 2) = 12 hours [1] (c) dC/dt = −12k e^(−kt), so at t = 2, dC/dt = −12 × ¼ ln 2 × e^(−½ ln 2) [1]; decreasing at 1.47 mg/L per hour (3 s.f.) [1] Examiner insight: in (b), using the rounded k = 0.173 still gives 12.0, but giving 12.0 without showing ln 8 / k risks the method mark.

11. (a) At t = 0, P = A, so A = 24 000 [1]; 31 000 = 24 000e^(10k) [1]; k = (1/10) ln(31/24) = 0.0256 (3 s.f.) [1] (b) t = 30 [1]; P = 24 000e^(30k) = 51 700 (3 s.f.) [1] (c) e^(kt) = 50 000/24 000, t = ln(25/12)/k = 28.7 [1]; so 2038 [1] (d) Limitation: the model grows without limit, but housing, land and services limit a town’s size, so long-term predictions are likely to be too high [1]. Refinement: use a model whose growth rate falls so the population levels off at a maximum, or refit k using newer census data [1] Examiner insight: in (c), t = 28.7 means 28.7 years after 2010, which is during 2038; answering 2039 or 28.7 alone loses the final mark.

Where marks are usually lost

• Treating log(a + b) as log a + log b when combining logs.
• Not checking that every log argument is positive after solving, so an invalid root is kept.
• In hidden quadratics, losing a coefficient (3^(2x + 1) is 3 × 9ˣ, not 9ˣ) or keeping a negative u.
• Missing brackets when bringing a power down: (x + 1) ln 5, not x + 1 ln 5.
• Giving decimals where “exact” is asked, or exact forms with no decimal where 3 s.f. is asked.
• Quoting the intercept of a log graph as the parameter itself (ln a instead of a).
• Using rounded intermediate values so early that the third significant figure changes.
• Answering a “which year” question with a number of years, or the wrong year.
• Stating a generic limitation (“models are not perfect”) with no link to the context.

Next steps

• Revision notes for this section
• Study guide for this section
• Differentiation practice for more on gradients of eᵏˣ and ln x
• AQA A-Level Mathematics hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA A-level Mathematics (7357) specification, version 1.3, 31 January 2018 (A-level exams June 2018 onwards), published by AQA. Section 3.7 F: Exponentials and logarithms, content F1 to F7.