Question Simulator

O Level Additional Mathematics (4037)

Logarithmic and exponential functions Banker

11 questions · about 5% of the marks we indexed · examined in 8 of 9 sessions.

Questions by year

2026

  • 2026 May/June · Paper 12 · Q9 (a, b)7 marksLaws of logarithms and change of base · Logarithmic and exponential equations

2025

  • 2025 Oct/Nov · Paper 12 · Q5 (a, b)7 marksLaws of logarithms and change of base · Logarithmic and exponential equations
  • 2025 Oct/Nov · Paper 12 · Q84 marksGraphs of exponential and log functions

2024

  • 2024 Oct/Nov · Paper 22 · Q5 (a, b)8 marksLogarithmic and exponential equations
  • 2024 May/June · Paper 12 · Q23 marksLaws of logarithms and change of base

2023

  • 2023 Oct/Nov · Paper 22 · Q4 (a, b)9 marksLogarithmic and exponential equations
  • 2023 May/June · Paper 12 · Q5 (b)6 marksLogarithmic and exponential equations

2022

  • 2022 Oct/Nov · Paper 22 · Q24 marksLogarithmic and exponential equations
  • 2022 Oct/Nov · Paper 22 · Q45 marksLogarithmic and exponential equations
  • 2022 May/June · Paper 12 · Q10 (a(i), a(ii), b)11 marksLaws of logarithms and change of base · Logarithmic and exponential equations
  • 2022 May/June · Paper 22 · Q5 (b)2 marksGraphs of exponential and log functions

References only, not question text. Open Official paper ↗ where Cambridge publishes the paper, or see the official Cambridge past-papers page; for other papers, ask your school. “Paper 42” means Paper 4, variant 2.

Practice questions

Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.

Q1Logarithmic and exponential equations · Core · Paper 2 style · 2 marks
  1. (a) Solve 3ˣ = 20, giving your answer correct to 3 significant figures. [2]
Show answer
  1. (a) x = 2.73
    x = lg 20/lg 3 (or ln 20/ln 3) M1; = 2.73 A1
Q2Logarithmic and exponential equations · Core · Paper 1 style · 3 marks
  1. (a) Do not use a calculator in this question.Solve log₂(x + 3) = 5. [2]
  2. (b) Solve log₅ y = −3, giving y as a fraction. [1]
Show answer
  1. (a) x = 29
    x + 3 = 2⁵ = 32 M1; x = 29 A1
  2. (b) y = 1/125
    y = 5⁻³ B1
Q3Logarithmic and exponential equations · Core · Paper 1 style · 3 marks
  1. (a) Do not use a calculator in this question.Solve lg(2x − 1) − lg(x − 4) = 1. [3]
Show answer
  1. (a) x = 39/8 (4.875)
    lg((2x − 1)/(x − 4)) = 1 M1; 2x − 1 = 10(x − 4) M1; 8x = 39, x = 39/8 A1
Q4Logarithmic and exponential equations · Core · Paper 2 style · 5 marks

A cup of tea in a Lahore café cools so that its temperature, T °C, t minutes after it is poured is given by T = 22 + 68e^(−0.04t).

  1. (a) Find the temperature of the tea when it is poured. [1]
  2. (b) Find the time taken for the tea to cool to 50 °C. Give your answer correct to 3 significant figures. [3]
  3. (c) State the temperature that the tea approaches as t becomes very large. [1]
Show answer
  1. (a) 90 °C
    22 + 68 = 90 B1
  2. (b) 22.2 minutes
    68e^(−0.04t) = 28 M1; −0.04t = ln(28/68) M1; t = 22.2 A1
  3. (c) 22 °C
    B1
Q5Logarithmic and exponential equations · Core · Paper 2 style · 2 marks
  1. (a) Solve e^(3x − 1) = 7, giving your answer correct to 3 significant figures. [2]
Show answer
  1. (a) x = 0.982
    3x − 1 = ln 7 M1; x = (1 + ln 7)/3 = 0.982 A1
Q6Logarithmic and exponential equations · Standard · Paper 1 style · 4 marks
  1. (a) Do not use a calculator in this question.Solve log₃ x + log₃(x − 6) = 3. [4]
Show answer
  1. (a) x = 9
    log₃(x(x − 6)) = 3 M1; x² − 6x − 27 = 0 A1; (x − 9)(x + 3) = 0 M1; x = 9 only, since x = −3 makes the logs undefined A1
Q7Logarithmic and exponential equations · Standard · Paper 2 style · 5 marks
  1. (a) Solve 5²ˣ − 7(5ˣ) + 12 = 0. Give your answers correct to 3 significant figures. [5]
Show answer
  1. (a) x = 0.683, 0.861
    Let u = 5ˣ: u² − 7u + 12 = 0 M1; u = 3, 4 A1; 5ˣ = 3 → x = lg 3/lg 5 = 0.683 M1 A1; 5ˣ = 4 → x = 0.861 A1
Q8Logarithmic and exponential equations · Standard · Paper 1 style · 4 marks
  1. (a) Do not use a calculator in this question.Solve e²ˣ − 5eˣ + 6 = 0, giving your answers in exact form. [4]
Show answer
  1. (a) x = ln 2, ln 3
    u² − 5u + 6 = 0 with u = eˣ M1; (u − 2)(u − 3) = 0 A1; x = ln 2 A1, x = ln 3 A1
Q9Logarithmic and exponential equations · Standard · Paper 2 style · 4 marks
  1. (a) Solve 2ˣ⁺¹ = 3ˣ⁻¹, giving your answer correct to 3 significant figures. [4]
Show answer
  1. (a) x = 4.42
    (x + 1)lg 2 = (x − 1)lg 3 M1; x(lg 3 − lg 2) = lg 3 + lg 2 M1 A1; x = lg 6/lg 1.5 = 4.42 A1
Q10Logarithmic and exponential equations · Standard · Paper 2 style · 6 marks

The number of residents, P, in a new housing scheme near Islamabad is modelled by P = 2400e^(0.035t), where t is the time in years after the scheme opened.

  1. (a) State the number of residents when the scheme opened. [1]
  2. (b) Find the number of residents after 8 years, to the nearest whole number. [2]
  3. (c) Find the value of t when P = 5000, giving your answer correct to 3 significant figures. [3]
Show answer
  1. (a) 2400
    B1
  2. (b) 3176
    2400e^0.28 M1; = 3176 A1
  3. (c) t = 21.0
    e^(0.035t) = 5000/2400 M1; 0.035t = ln(25/12) M1; t = 20.97… = 21.0 A1
Q11Logarithmic and exponential equations · Standard · Paper 1 style · 3 marks
  1. (a) Do not use a calculator in this question.Solve ln(3x + 1) − ln(x − 1) = ln 5. [3]
Show answer
  1. (a) x = 3
    ln((3x + 1)/(x − 1)) = ln 5 M1; 3x + 1 = 5x − 5 M1; x = 3 A1
Q12Logarithmic and exponential equations · Standard · Paper 2 style · 4 marks
  1. (a) Solve 4ˣ = 3(2ˣ) + 10, giving your answer correct to 3 significant figures. [4]
Show answer
  1. (a) x = 2.32
    u = 2ˣ: u² − 3u − 10 = 0 M1; (u − 5)(u + 2) = 0 A1; 2ˣ = −2 has no solution B1; x = lg 5/lg 2 = 2.32 A1
Q13Logarithmic and exponential equations · Standard · Paper 2 style · 4 marks
  1. (a) Solve 3e²ˣ + eˣ − 2 = 0. Give your answer in exact form and also correct to 3 significant figures. [4]
Show answer
  1. (a) x = ln(2/3) = −0.405
    (3eˣ − 2)(eˣ + 1) = 0 M1 A1; eˣ = −1 has no solution B1; x = ln(2/3) = −0.405 A1
Q14Logarithmic and exponential equations · Standard · Paper 1 style · 4 marks
  1. (a) Do not use a calculator in this question.By changing log₄(3x + 4) to base 2, solve log₂ x = log₄(3x + 4). [4]
Show answer
  1. (a) x = 4
    log₄(3x + 4) = ½ log₂(3x + 4) M1; 2 log₂ x = log₂(3x + 4) → x² = 3x + 4 M1 A1; x = 4 (x = −1 rejected) A1
Q15Logarithmic and exponential equations · Challenge · Paper 1 style · 5 marks
  1. (a) Do not use a calculator in this question.By writing logₓ 4 in terms of log₄ x, solve log₄ x + logₓ 4 = 5/2. [5]
Show answer
  1. (a) x = 16, x = 2
    logₓ 4 = 1/log₄ x M1; t + 1/t = 5/2 → 2t² − 5t + 2 = 0 M1 A1; t = 2, ½ A1; x = 16, 2 A1
Q16Logarithmic and exponential equations · Challenge · Paper 1 style · 5 marks
  1. (a) Do not use a calculator in this question.Solve 2 log₃ x − log₃(x + 4) = 1, giving your answer in exact form. [5]
Show answer
  1. (a) x = (3 + √57)/2
    log₃(x²/(x + 4)) = 1 M1 M1; x² = 3(x + 4) → x² − 3x − 12 = 0 A1; x = (3 ± √57)/2 M1; reject the negative root since log₃ x needs x > 0 A1
Q17Logarithmic and exponential equations · Challenge · Paper 2 style · 6 marks

The mass, m grams, of a radioactive substance in a hospital laboratory is given by m = 80e^(−kt), where t is the time in days and k is a positive constant. After 12 days the mass is 40 g.

  1. (a) Show that k = (ln 2)/12. [2]
  2. (b) Find the mass after 30 days, correct to 3 significant figures. [2]
  3. (c) Find the time taken for the mass to fall to 5 g. [2]
Show answer
  1. (a) Shown
    40 = 80e^(−12k) → e^(−12k) = ½ M1; −12k = ln ½ = −ln 2 A1
  2. (b) 14.1 g
    80e^(−30 ln2/12) = 80 × 2^(−2.5) M1; = 14.1 A1
  3. (c) 48 days
    e^(−kt) = 1/16 M1; t = 12 ln 16/ln 2 = 48 A1
Q18Logarithmic and exponential equations · Challenge · Paper 2 style · 5 marks
  1. (a) Solve the simultaneous equationslg x + lg y = 2,x − y = 15. [5]
Show answer
  1. (a) x = 20, y = 5
    lg(xy) = 2 → xy = 100 M1 A1; (y + 15)y = 100 → y² + 15y − 100 = 0 M1; y = 5 (y = −20 rejected, logs undefined) A1; x = 20 A1
Q19Logarithmic and exponential equations · Challenge · Paper 1 style · 5 marks
  1. (a) Do not use a calculator in this question.Solve the simultaneous equations2ˣ × 4ʸ = 32,3ˣ ÷ 9ʸ = 1/27. [5]
Show answer
  1. (a) x = 1, y = 2
    2ˣ⁺²ʸ = 2⁵ → x + 2y = 5 M1 A1; 3ˣ⁻²ʸ = 3⁻³ → x − 2y = −3 M1 A1; x = 1, y = 2 A1
Q20Logarithmic and exponential equations · Challenge · Paper 2 style · 5 marks
  1. (a) Solve ln x + ln(x + 2) = 1. Give your answer correct to 3 significant figures. [5]
Show answer
  1. (a) x = 0.928
    ln(x(x + 2)) = 1 M1; x² + 2x − e = 0 A1; x = (−2 ± √(4 + 4e))/2 M1; x = −1 + √(1 + e) A1; = 0.928 (negative root rejected) A1
Q21Laws of logarithms and change of base · Core · Paper 1 style · 3 marks
  1. (a) Do not use a calculator in this question.Show that 2 lg 5 + lg 12 − lg 3 = 2. [3]
Show answer
  1. (a) Shown
    2 lg 5 = lg 25 B1; lg(25 × 12 ÷ 3) M1; = lg 100 = 2 A1
Q22Laws of logarithms and change of base · Core · Paper 1 style · 6 marks

It is given that logₐ 2 = p and logₐ 3 = q.

  1. (a) Express logₐ 18 in terms of p and q. [2]
  2. (b) Express logₐ(8/9) in terms of p and q. [2]
  3. (c) Express logₐ(a√6) in terms of p and q. [2]
Show answer
  1. (a) p + 2q
    18 = 2 × 3² M1 A1
  2. (b) 3p − 2q
    B2 (B1 for one term)
  3. (c) 1 + ½(p + q)
    logₐ a = 1 B1; ½ logₐ 6 B1
Q23Laws of logarithms and change of base · Standard · Paper 1 style · 4 marks
  1. (a) Use the change of base rule to show that log₉ x = ½ log₃ x. [2]
  2. (b) Do not use a calculator in this part.Hence solve log₃ x + log₉ x = 6. [2]
Show answer
  1. (a) Shown
    log₉ x = log₃ x/log₃ 9 M1; log₃ 9 = 2 A1
  2. (b) x = 81
    (3/2) log₃ x = 6 → log₃ x = 4 M1; x = 81 A1
Q24Laws of logarithms and change of base · Standard · Paper 2 style · 4 marks
  1. (a) Use the change of base rule to evaluate log₇ 50, giving your answer correct to 3 significant figures. [2]
  2. (b) Evaluate log₀.₅ 10 correct to 3 significant figures, and explain why the answer is negative. [2]
Show answer
  1. (a) 2.01
    lg 50/lg 7 M1; = 2.01 A1
  2. (b) −3.32; negative because 0.5 raised to a positive power is less than 1, so a power giving 10 must be negative
    lg 10/lg 0.5 = −3.32 B1; reason B1
Q25Laws of logarithms and change of base · Challenge · Paper 1 style · 3 marks
  1. (a) Do not use a calculator in this question.Find the exact value of (log₂ 27)(log₃ 16). [3]
Show answer
  1. (a) 12
    log₂ 27 = 3 log₂ 3, log₃ 16 = 4 log₃ 2 M1; log₂ 3 × log₃ 2 = 1 (change of base) M1; 3 × 4 = 12 A1
Q26Laws of logarithms and change of base · Challenge · Paper 1 style · 6 marks

It is given that lg x = a and lg y = b.

  1. (a) Express lg(100x³/√y) in terms of a and b. [3]
  2. (b) Express log₁₀₀(xy) in terms of a and b. [2]
  3. (c) Express logₓ y in terms of a and b. [1]
Show answer
  1. (a) 2 + 3a − ½b
    lg 100 = 2 B1; 3a B1; −½b B1
  2. (b) (a + b)/2
    lg(xy)/lg 100 M1; = (a + b)/2 A1
  3. (c) b/a
    B1
Q27Graphs of exponential and log functions · Core · Paper 1 style · 4 marks
  1. (a) For the curve y = e²ˣ − 3, find the exact coordinates of the points where it meets the axes. [3]
  2. (b) State the equation of the asymptote of the curve. [1]
Show answer
  1. (a) (0, −2) and (½ ln 3, 0)
    x = 0: y = 1 − 3 = −2 B1; e²ˣ = 3 M1; x = ½ ln 3 A1
  2. (b) y = −3
    B1
Q28Graphs of exponential and log functions · Standard · Paper 1 style · 5 marks
  1. (a) For the curve y = ln(x + 2) + 1, state the equation of the asymptote. [1]
  2. (b) Find the exact coordinates of the points where the curve meets the axes. [3]
  3. (c) Describe the shape of the curve. [1]
Show answer
  1. (a) x = −2
    B1
  2. (b) (0, 1 + ln 2) and (e⁻¹ − 2, 0)
    x = 0: y = ln 2 + 1 B1; ln(x + 2) = −1 M1; x = e⁻¹ − 2 A1
  3. (c) Increasing for all x > −2, with gradient decreasing; it approaches x = −2 downwards
    B1
Q29Graphs of exponential and log functions · Standard · Paper 2 style · 6 marks
  1. (a) The curve y = 3e⁻ˣ + 1 is drawn for all real x. State the coordinates of the y-intercept, the equation of the asymptote and the range of y. [3]
  2. (b) By considering the graphs of y = 3e⁻ˣ + 1 and y = x, explain why the equation 3e⁻ˣ + 1 = x has exactly one real root, and state an interval of the form n < x < n + 1 (n an integer) in which it lies. [3]
Show answer
  1. (a) (0, 4); y = 1; y > 1
    B1 B1 B1
  2. (b) One root, between x = 1 and x = 2
    y = 3e⁻ˣ + 1 decreases, y = x increases, so they cross once B1; at x = 1: 3e⁻¹ + 1 = 2.10 > 1; at x = 2: 3e⁻² + 1 = 1.41 < 2 M1; root in 1 < x < 2 A1
Q30Graphs of exponential and log functions · Challenge · Paper 2 style · 6 marks

The curve C has equation y = 2 ln(3x − 4), for x > 4/3.

  1. (a) State the equation of the asymptote of C. [1]
  2. (b) Find the coordinates of the point where C crosses the x-axis. [2]
  3. (c) Find the value of x for which y = 4, giving your answer in exact form and correct to 3 significant figures. [3]
Show answer
  1. (a) x = 4/3
    B1
  2. (b) (5/3, 0)
    3x − 4 = 1 M1; x = 5/3 A1
  3. (c) x = (e² + 4)/3 = 3.80
    ln(3x − 4) = 2 M1; 3x − 4 = e² M1; x = (e² + 4)/3 = 3.80 A1

Similar questions

Questions that test the same subtopic in different years, grouped so you can work on one skill at a time.

References only: find each question in its paper (official link where available, otherwise through your school).

Logarithmic and exponential equations 8 questions · 7/9 sessions
Laws of logarithms and change of base 4 questions · 4/9 sessions
Graphs of exponential and log functions Rare2 questions · 2/9 sessions

Subtopics

Our analysis of how often each subtopic was examined and how many marks it carried, 2022 to May/June 2026.

SubtopicMarks indexedSessions
Logarithmic and exponential equations 477/9
Laws of logarithms and change of base 134/9
Graphs of exponential and log functions Rare62/9

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