Mathematics Advanced • Year 11 • Module 4 • Lesson 6

Graphs of Logarithmic Functions

Apply log-graph features to real contexts: sound, earthquakes, finance, pH, and the inverse-graph principle.

Apply · Problem Set

Problem 1 — Decibel scale (sound intensity)

The loudness L (in decibels) of a sound is related to its intensity I (in W/m²) by

L(I) = 10 log₁₀(I / I₀), where I₀ = 10⁻¹² W/m²

Set up: What are we solving for?

(i) Treating L as a function of x = I/I₀, state the vertical asymptote and the x-intercept of L(x) = 10 log₁₀(x), and explain the physical meaning of each. 3 marks

(ii) A whisper has I = 10⁻¹⁰ W/m². Compute L. 1 mark

(iii) A jet engine measures 130 dB. How many times more intense is the jet than the whisper? Use logarithm properties to argue this without a calculator. 3 marks

Stuck? Revisit lesson § Concept — log graphs convert multiplicative leaps into additive steps.

Problem 2 — Earthquake magnitudes (Richter scale)

The Richter magnitude of an earthquake of seismic-wave amplitude A is

M(A) = log₁₀(A / A₀), A₀ = reference amplitude

Set up: What are we solving for?

(i) Treating M as a function of x = A/A₀, sketch the shape of M(x) = log₁₀(x) for 0 < x ≤ 1000, marking the asymptote and the x-intercept. 2 marks

(ii) A magnitude-6 earthquake corresponds to x = ____________ (give a power of 10). A magnitude-7 earthquake corresponds to x = ____________. By how many multiplicative factors do the amplitudes differ? 2 marks

(iii) Explain in one sentence why M(0) is undefined and what this means physically about an earthquake with zero amplitude. 2 marks

Problem 3 — Doubling-time curve (compound growth)

An investment growing at annual rate r has a doubling time given (to a close approximation) by

T(r) = ln 2 / r, r > 0 (decimal)

For this problem, write T as a function of an alternative variable u = r/(ln 2), so T(u) = 1/u, and contrast that with a related log curve y = −log₂(u).

Set up: What are we solving for?

(i) Sketch on the same axes for u > 0: y = log₂(u) and y = −log₂(u). State the asymptote, the common x-intercept, and the line of symmetry between the two curves. 3 marks

(ii) Explain in one sentence why y = −log₂(u) is the reflection of y = log₂(u) in the x-axis, and state what happens to the asymptote and to the x-intercept under this reflection. 2 marks

(iii) The "Rule of 72" approximates T(r) ≈ 72 / (100r). Use the exact T(r) = ln 2 / r to compute T for r = 0.06 (6%) and compare with the Rule-of-72 estimate. 2 marks

Stuck on (ii)? Revisit lesson § Trap 01 and § Concept — negative outside the log reflects across the x-axis.

Problem 4 — pH scale (chemistry)

The pH of an aqueous solution is

pH(c) = −log₁₀(c), c = hydrogen-ion concentration in mol/L, c > 0

Set up: What are we solving for?

(i) State the vertical asymptote of pH(c) and explain in one sentence why this asymptote is never reached by any real solution. 2 marks

(ii) Pure water has c = 10⁻⁷ mol/L. Compute pH(10⁻⁷) and identify which key feature of the basic graph y = log₁₀(c) gives you this answer immediately. 2 marks

(iii) A lemon has pH ≈ 2. A glass of black coffee has pH ≈ 5. Without computing c for each, use the graph's behaviour to state how many times more concentrated the lemon is in H⁺ than the coffee. Justify in one sentence. 2 marks

Problem 5 — Inverse-graph reflection

The exponential y = e^{x − 1} has inverse y = ln(x) + 1. The two curves are reflections of each other in the line y = x.

Set up: What are we solving for?

(i) Verify algebraically that the point (1, 1) lies on both curves, and state why this point is fixed under reflection in y = x. 2 marks

(ii) The point (0, e⁻¹) lies on the exponential. Use the reflection principle to write down a corresponding point on the logarithmic curve, and verify that it satisfies y = ln(x) + 1. 2 marks

(iii) State the asymptote of each curve, explain how the two asymptotes correspond under reflection in y = x, and explain in one sentence why a vertical asymptote of one becomes a horizontal asymptote of the other. 3 marks

Stuck? Revisit lesson § Key terms — Inverse graph.

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Problem 1 — Decibels

Set up. We treat L as a log function of the intensity ratio x = I/I₀, then use the standard features of a log graph to interpret the scale.

(i) Asymptote: x = 0 — corresponds to I = 0 (no sound), where L → −∞ (the scale is undefined). x-intercept: at x = 1, log₁₀(1) = 0, so L = 0 dB. Physically, L = 0 dB is the threshold of human hearing (I = I₀).

(ii) L = 10 log₁₀(10⁻¹⁰ / 10⁻¹²) = 10 log₁₀(10²) = 10 × 2 = 20 dB.

(iii) The jet at 130 dB has 10 log₁₀(x_jet) = 130, so x_jet = 10¹³. The whisper has x_w = 10². Ratio = 10¹³/10² = 10¹¹ — the jet is 10¹¹ (a hundred billion) times more intense. The decibel scale compresses this enormous ratio (a 110-dB difference) into a comprehensible number — the value of the log scale.

Problem 2 — Richter

Set up. We are using the basic shape of y = log₁₀(x) to interpret magnitudes as exponents of the amplitude ratio.

(i) Asymptote x = 0; x-intercept (1, 0). Curve rises slowly from very negative values near x = 0, crosses the x-axis at x = 1, and at x = 10 reaches M = 1, x = 100 reaches M = 2, x = 1000 reaches M = 3.

(ii) Magnitude 6 ⇒ x = 10⁶. Magnitude 7 ⇒ x = 10⁷. Multiplicative factor: 10⁷/10⁶ = 10. Every whole-number step on the Richter scale is a tenfold increase in amplitude.

(iii) M(0) = log₁₀(0) is undefined because 0 is not in the domain of any logarithm (10^anything > 0, so no power of 10 equals 0). Physically, "zero amplitude" means no earthquake at all, which is outside the model.

Problem 3 — Doubling-time and reflection in x-axis

Set up. We are comparing y = log₂(u) and y = −log₂(u) to see what a leading minus does to a log graph.

(i) Both curves: asymptote x = 0; common x-intercept (1, 0). y = log₂(u) is increasing (passes through (2, 1), (4, 2)); y = −log₂(u) is decreasing (passes through (2, −1), (4, −2)). The two curves are mirror images across the x-axis.

(ii) A leading minus sends every y-value to its negative, which is exactly reflection in the x-axis. The asymptote x = 0 is on the x-axis (vertical line) so it maps to itself; the x-intercept (1, 0) is on the x-axis so it is fixed.

(iii) Exact: T(0.06) = ln 2 / 0.06 ≈ 0.693/0.06 ≈ 11.55 years. Rule-of-72: 72/6 = 12 years. The Rule-of-72 over-estimates by ≈ 0.45 yr (≈ 3.8%) — close enough for mental arithmetic.

Problem 4 — pH scale

Set up. We are reading off features of pH(c) = −log₁₀(c), which is a reflection of log₁₀(c) in the x-axis.

(i) Asymptote c = 0. No real aqueous solution has exactly zero H⁺ concentration (water auto-ionises), so c > 0 always, and the asymptote is approached but never reached.

(ii) pH(10⁻⁷) = −log₁₀(10⁻⁷) = −(−7) = 7. Feature used: log₁₀(10^k) = k for any integer k — a direct read-off from the graph's "power-of-10 stair steps".

(iii) Difference in pH = 5 − 2 = 3 units of −log₁₀, so c ratio = 10³. The lemon is 1000 times more concentrated in H⁺ than the coffee. The leading minus reverses the direction (lower pH ⇒ higher c), but each unit step is still a factor of 10.

Problem 5 — Inverse-graph reflection

Set up. We are using the inverse-function principle: exponential and logarithm are reflections in y = x.

(i) On y = e^{x − 1}: at x = 1, y = e⁰ = 1. ✓ On y = ln(x) + 1: at x = 1, y = ln(1) + 1 = 0 + 1 = 1. ✓ (1, 1) lies on y = x, so reflection in y = x maps it to itself — it is a fixed point of the reflection.

(ii) Reflection in y = x swaps coordinates, so (0, e⁻¹) maps to (e⁻¹, 0). Verify: y = ln(e⁻¹) + 1 = −1 + 1 = 0. ✓

(iii) Exponential y = e^{x − 1}: horizontal asymptote y = 0. Logarithm y = ln(x) + 1: vertical asymptote x = 0. The line y = 0 (x-axis) reflects to x = 0 (y-axis) across y = x: every horizontal asymptote of an exponential becomes a vertical asymptote of its inverse, because "approaches y = 0 as x → −∞" becomes "approaches x = 0 as y → −∞" after swapping the roles of x and y.