Practice Maths

Topic Review — Logarithms and Logarithmic Functions — Solutions

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★ U2T2 — Logarithms and Logarithmic Functions Review

This review covers all three lessons in this topic. Attempt all questions. Click ▶ Show Answer to reveal each solution.

Covers: Evaluating logarithms • Logarithmic laws • Converting between forms • Graphs (asymptote, intercepts, domain, transformations) • Finding equations from graphs • Solving logarithmic equations • Applications (pH, Richter scale, population growth, compound interest)

  1. Fluency

    Q1 — Evaluating logarithms

    Evaluate each logarithm without a calculator.

    (a) log2(32)    (b) log3(1/9)    (c) log(10 000)    (d) log5(57)    (e) ln(e3)    (f) log4(8)

    (a) log2(32) = log2(25) = 5

    (b) log3(1/9) = log3(3−2) = −2

    (c) log(10 000) = log(104) = 4

    (d) log5(57) = 7 (log and exponential with same base cancel)

    (e) ln(e3) = 3

    (f) log4(8) = log4(43/2) = 3/2   [since 43/2 = (22)3/2 = 23 = 8]

  2. Fluency

    Q2 — Converting between exponential and logarithmic form

    Convert each to the other form.

    (a) 28 = 256    (b) 10−4 = 0.0001    (c) log7(343) = 3    (d) ln(x) = k

    (a) 28 = 256 → log2(256) = 8

    (b) 10−4 = 0.0001 → log(0.0001) = −4

    (c) log7(343) = 3 → 73 = 343

    (d) ln(x) = k → ek = x

  3. Fluency

    Q3 — Applying logarithmic laws

    Write each expression as a single logarithm in simplest form.

    (a) log2(5) + log2(8)    (b) 3log(x) − log(y)    (c) 2log3(6) − log3(4)    (d) ½log(x4y2)

    (a) log2(5 × 8) = log2(40)   [cannot simplify further since 40 is not a power of 2]   log2(40)

    (b) log(x3) − log(y) = log(x3/y)

    (c) log3(62) − log3(4) = log3(36/4) = log3(9) = log3(32) = 2

    (d) ½log(x4y2) = ½[log(x4) + log(y2)] = ½[4log(x) + 2log(y)] = 2log(x) + log(y)   or equivalently   log(x2y)

  4. Fluency

    Q4 — Key features of a logarithmic graph

    For the function y = 3log2(x − 2) − 6, state:

    (a) the vertical asymptote    (b) the domain    (c) the x-intercept    (d) the y-intercept (if it exists)    (e) whether the function is increasing or decreasing

    (a) Vertical asymptote: x = 2

    (b) Domain: x − 2 > 0 → x > 2

    (c) x-intercept (set y = 0): 3log2(x−2) = 6 → log2(x−2) = 2 → x−2 = 4 → x = 6. x-intercept: (6, 0)

    (d) Domain is x > 2, so x = 0 is not in the domain. No y-intercept.

    (e) A = 3 > 0 and base 2 > 1, so the function is increasing.

  5. Understanding

    Q5 — Describing and applying transformations

    Starting from y = log3(x), describe the transformations and state the new asymptote and domain for each function.

    (a) y = log3(x + 5) − 2    (b) y = −2log3(x − 1) + 4

    (a) y = log3(x + 5) − 2:
    Transformations: horizontal shift left 5 units (h = −5) and vertical shift down 2 units.
    Vertical asymptote: x = −5.   Domain: x > −5.

    (b) y = −2log3(x − 1) + 4:
    Transformations: horizontal shift right 1 unit, vertical stretch by factor 2, reflection in the x-axis (because A = −2), and vertical shift up 4 units.
    Vertical asymptote: x = 1.   Domain: x > 1.   Function is decreasing.

  6. Understanding

    Q6 — Finding the equation from a graph

    A logarithmic function of the form y = A log(x) + k passes through the points (1, 5) and (100, −1). Find the values of A and k, and write the equation.

    At (1, 5): A log(1) + k = A(0) + k = k = 5. So k = 5.

    At (100, −1): A log(100) + 5 = A(2) + 5 = −1 → 2A = −6 → A = −3. So A = −3.

    Equation: y = −3log(x) + 5

    This is a decreasing function (A < 0), which makes sense as larger x gives smaller y.

  7. Understanding

    Q7 — Solving logarithmic equations

    Solve each equation. Show all working and check your solutions.

    (a) log4(3x + 1) = 3    (b) log(x2) = 4    (c) log2(x + 3) = log2(5x − 9)

    (a) log4(3x+1) = 3 → 3x+1 = 43 = 64 → 3x = 63 → x = 21. Check: 3(21)+1=64>0 ✓

    (b) log(x2) = 4 → x2 = 104 = 10 000 → x = ±100. Domain: x2 > 0 for all x ≠ 0. Both x = 100 and x = −100 give x2 = 10 000 > 0 ✓. x = 100 or x = −100

    (c) log2(x+3) = log2(5x−9) → x+3 = 5x−9 → 12 = 4x → x = 3. Check: x+3=6>0 ✓ and 5(3)−9=6>0 ✓. x = 3

  8. Understanding

    Q8 — Solving using log laws first

    Solve log3(x + 6) + log3(x) = 3. Show all working and state any rejected solutions.

    Product law: log3(x(x + 6)) = 3

    Convert: x(x + 6) = 33 = 27

    x2 + 6x − 27 = 0 → (x + 9)(x − 3) = 0 → x = −9 or x = 3

    Domain check: need x > 0 AND x + 6 > 0, so x > 0.

    x = −9: fails (x must be positive). x = 3: ✓

    x = 3   (x = −9 is extraneous)

  9. Understanding

    Q9 — pH application

    A chemist measures the pH of two solutions:

    Solution A has pH = 3.4    Solution B has pH = 6.2

    (a) Find the [H+] concentration of each solution. Give answers in scientific notation to 2 significant figures.
    (b) How many times greater is the [H+] concentration in Solution A than in Solution B?

    (a) [H+]A = 10−3.4 = 10−4 × 100.6 = 10−4 × 3.981 ≈ 4.0 × 10−4 mol/L

    [H+]B = 10−6.2 = 10−7 × 100.8 = 10−7 × 6.310 ≈ 6.3 × 10−7 mol/L

    (b) Ratio = [H+]A / [H+]B = 10−3.4 / 10−6.2 = 106.2 − 3.4 = 102.8 ≈ 631.
    Solution A has a [H+] concentration approximately 631 times greater than Solution B.

  10. Understanding

    Q10 — Exponential equation solved using logarithms

    Solve each equation. Give exact answers and decimal approximations (3 d.p.).

    (a) 4x = 100    (b) 2 e3x = 50    (c) 52x − 1 = 30

    (a) 4x = 100 → x log(4) = log(100) = 2 → x = 2/log(4) = 2/0.60206.   Exact: x = 2/log(4)   Decimal: x ≈ 3.322

    (b) 2 e3x = 50 → e3x = 25 → 3x = ln(25) → x = ln(25)/3.   Exact: x = ln(25)/3   Decimal: x = 3.2189/3 ≈ 1.073

    (c) 52x − 1 = 30 → (2x−1)log(5) = log(30) → 2x−1 = log(30)/log(5) = 1.47712/0.69897 ≈ 2.1133 → 2x ≈ 3.1133 → x ≈ 1.5567.
    Exact: x = (1 + log(30)/log(5)) / 2 = (1 + log5(30)) / 2   Decimal: x ≈ 1.557

  11. Understanding

    Q11 — Richter scale comparison

    The 1906 San Francisco earthquake had magnitude M = 7.9. The 1989 Loma Prieta earthquake had M = 6.9.

    (a) How many times more intense was the 1906 earthquake than the 1989 earthquake?
    (b) An aftershock was 250 times less intense than the 1989 earthquake. Find its magnitude, correct to 1 decimal place.

    (a) Ratio = 107.9 − 6.9 = 101 = 10 times more intense

    (b) Intensity of aftershock = I1989 / 250.
    Maftershock = log(Iaftershock / I0) = log(I1989 / (250 I0)) = log(I1989/I0) − log(250) = 6.9 − log(250)
    log(250) = log(1000/4) = 3 − log(4) = 3 − 0.6021 = 2.3979
    Maftershock = 6.9 − 2.3979 ≈ 4.5

  12. Problem Solving

    Q12 — Doubling time and half-life

    Challenge. A bacterial culture grows according to N = N0 e0.35t, where t is time in hours.

    (a) Find the doubling time (the time for the population to double), correct to 2 decimal places.
    (b) A radioactive isotope decays according to M = M0 e−0.0231t. Find its half-life in years, correct to 1 decimal place.
    (c) If the half-life of the isotope is approximately 30 years, what is the decay rate constant k (4 d.p.)?

    (a) Set N = 2N0:   e0.35t = 2 → 0.35t = ln(2) → t = ln(2)/0.35 = 0.6931/0.35 ≈ 1.98 hours

    (b) Set M = M0/2:   e−0.0231t = 1/2 → −0.0231t = ln(1/2) = −ln(2) → t = ln(2)/0.0231 = 0.6931/0.0231 ≈ 30.0 years

    (c) At half-life t = 30: M0/2 = M0 e−30k → −30k = ln(1/2) = −ln(2) → k = ln(2)/30 = 0.6931/30 ≈ 0.0231

  13. Problem Solving

    Q13 — Finding the equation of a log graph from its sketch

    Challenge. A logarithmic function has the form y = A logb(x − h). Its graph has vertical asymptote x = 3, passes through (4, 0) and (12, 2).

    (a) Use the point (4, 0) to find h and explain why this fixes h.
    (b) Use the point (12, 2) along with your value of h to find A and b.
    (c) Write the complete equation.

    (a) The vertical asymptote is x = 3, so h = 3. Form: y = A logb(x − 3).
    At (4, 0): A logb(4 − 3) = A logb(1) = A(0) = 0 ✓. This is true for any A and b, so it confirms h = 3 but does not determine A or b.

    (b) At (12, 2): A logb(12 − 3) = A logb(9) = 2.
    We need more information to separate A and b. With one equation we can choose one freely. If we assume integer base: try b = 3: log3(9) = 2, so A(2) = 2 → A = 1. Let’s verify: y = log3(x − 3).
    At (4,0): log3(1)=0 ✓. At (12,2): log3(9)=2 ✓. A = 1, b = 3.

    (c) y = log3(x − 3)

  14. Problem Solving

    Q14 — Solving a logarithmic inequality

    Challenge. Solve the inequality log2(x + 1) + log2(x − 1) < 3. State the solution as an interval.

    Step 1 — Domain: Need x+1>0 AND x−1>0, so domain is x > 1.

    Step 2 — Combine: log2((x+1)(x−1)) < 3 → log2(x2−1) < 3

    Step 3 — Convert (base 2 > 1, so inequality direction is preserved): x2 − 1 < 23 = 8 → x2 < 9 → −3 < x < 3

    Step 4 — Intersect with domain: Domain requires x > 1. Intersect with −3 < x < 3: solution is 1 < x < 3, i.e. the interval (1, 3).

  15. Problem Solving

    Q15 — Multi-step application: decibels and intensity

    Challenge. The loudness of sound in decibels is L = 10log(I / I0), where I0 = 10−12 W/m2.

    (a) A rock concert has a loudness of 110 dB. Find its intensity I in W/m2.
    (b) A library has a loudness of 40 dB. How many times more intense is the concert than the library?
    (c) Two sounds have intensities I1 and I2. Show that the difference in their loudness levels is L1 − L2 = 10log(I1/I2).
    (d) By how many decibels must you increase a sound to triple its intensity?

    (a) 110 = 10log(I/10−12) → log(I/10−12) = 11 → I/10−12 = 1011 → I = 1011 × 10−12 = 0.1 W/m2

    (b) Library: L = 40 → Ilib = 10−12 × 104 = 10−8 W/m2.
    Ratio = Iconcert/Ilib = 0.1/10−8 = 10−1/10−8 = 107.
    The concert is 10 000 000 (107) times more intense than the library.   [Alternatively: difference = 110−40 = 70 dB, ratio = 107.]

    (c) L1 − L2 = 10log(I1/I0) − 10log(I2/I0) = 10[log(I1/I0) − log(I2/I0)]
    = 10log((I1/I0) / (I2/I0)) = 10log(I1/I2) ✓

    (d) Set I1/I2 = 3: ΔL = 10log(3) = 10 × 0.4771 ≈ 4.77 dB ≈ 4.8 dB