Mathematics Advanced • Year 11 • Module 4 • Lesson 5

Introduction to Logarithmic Functions

Apply logarithms to scientific scales (pH, decibels, Richter), to inverse questions, and to graph features.

Apply · Problem Set

Problem 1 — pH scale (chemistry context)

The pH of a solution with hydrogen-ion concentration [H+] (mol/L) is defined by:

pH = − log10 [H+]

Set up: What are we solving for?

(i) Compute the pH of a solution with [H+] = 1 × 10−3 mol/L.   1 mark

(ii) A second solution has pH = 5. Find its [H+] in mol/L.   2 marks

(iii) Explain in one sentence why a one-unit drop in pH (e.g. from 7 to 6) corresponds to a ten-fold increase in [H+].   2 marks

Stuck? pH is a base-10 log, so each unit is a factor of 10.

Problem 2 — Richter magnitude (geology)

The Richter magnitude M of an earthquake whose largest seismic-wave amplitude is A (relative to a calibration reference) is given by M = log10 A.

Set up: What are we solving for?

(i) Earthquake X has A = 10 000. Find its magnitude MX.   1 mark

(ii) Earthquake Y has magnitude 6.5. Find its amplitude AY.   2 marks

(iii) Quake Z has magnitude 8 and quake W has magnitude 5. How many times stronger is Z's amplitude than W's? Show one line of log arithmetic.   2 marks

Problem 3 — Decibels (acoustics)

Sound intensity I (in W/m2) is related to a perceptible loudness L (in decibels, dB) by

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

Set up: What are we solving for?

(i) A whisper has I = 10−10 W/m2. Compute L.   2 marks

(ii) A rock concert has L = 110 dB. Find I (in W/m2) in scientific notation.   2 marks

(iii) By how many decibels does the loudness change if the intensity doubles? Give an exact answer in terms of log10 2 and to 1 dp.   2 marks

Stuck on (iii)? ΔL = 10 log10 (2I / I) = 10 log10 2.

Problem 4 — Logs and exponentials as inverses

Let f(x) = 2x and g(x) = log2 x.

Set up: What are we solving for?

(i) Verify by computation that f(g(x)) = x and g(f(x)) = x for x in the appropriate domain.   2 marks

(ii) State the domain of f, the domain of g, and the range of each. Explain in one sentence how these swap under the inverse relationship.   3 marks

(iii) The graph of g is the reflection of the graph of f in which line? Justify in one line.   1 mark

Problem 5 — Reading features of y = log2 x

Consider y = log2 x.

Set up: What are we solving for?

(i) State the domain, the vertical asymptote, and the x-intercept.   2 marks

(ii) Compute log2 2, log2 4, log2 8, and log2 (1/2). Plot these four points (sketch) and join them with a smooth log curve.   3 marks

(iii) Describe the behaviour of y as x → 0+ and as x → ∞ in one sentence each.   2 marks

Stuck on the asymptote? Loga x is undefined at x = 0; what does this imply geometrically?

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

Problem 1 — pH

Set up. Use pH = −log10 [H+] in both directions and reason about scale.

(i) pH = −log10 (10−3) = −(−3) = 3.

(ii) 5 = −log10 [H+] ⇒ log10 [H+] = −5 ⇒ [H+] = 10−5 mol/L.

(iii) Because pH = −log10 [H+], dropping pH by 1 means log10 [H+] increases by 1, so [H+] is multiplied by 101 = 10 (a ten-fold increase).

Problem 2 — Richter

Set up. Use M = log10 A both ways and compute amplitude ratios.

(i) MX = log10 10 000 = 4.

(ii) 6.5 = log10 AY ⇒ AY = 106.53.16 × 106.

(iii) MZ − MW = 8 − 5 = 3 = log10 (AZ / AW) ⇒ AZ / AW = 103 = 1000 times stronger.

Problem 3 — Decibels

Set up. Use L = 10 log10 (I / I0) in both directions and find ΔL for intensity-doubling.

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

(ii) 110 = 10 log10 (I / I0) ⇒ 11 = log10 (I / I0) ⇒ I / I0 = 1011 ⇒ I = 1011 × 10−12 = 10−1 = 0.1 W/m2.

(iii) ΔL = 10 log10 (2I / I) = 10 log10 2 (exact) ≈ 10 × 0.3010 ≈ 3.0 dB. (A standard result: doubling intensity adds ≈ 3 dB.)

Problem 4 — Inverses

Set up. Verify composition gives x; describe domain/range swap and graphical symmetry.

(i) f(g(x)) = 2log2 x = x (for x > 0).   g(f(x)) = log2 2x = x (for all real x). ✓

(ii) Domain of f = ℝ (all reals); range of f = (0, ∞).   Domain of g = (0, ∞); range of g = ℝ. Under the inverse relationship the domain of one function becomes the range of the other and vice versa — i.e. the (x, y) pairs of f are reversed to (y, x) for g.

(iii) Reflection in the line y = x. Reason: every inverse function's graph is the reflection of the original's graph in y = x, because swapping (x, y) ↔ (y, x) is exactly that reflection.

Problem 5 — Features of y = log2 x

Set up. Identify domain, asymptote and key points, then sketch and describe limiting behaviour.

(i) Domain: x > 0, i.e. (0, ∞). Vertical asymptote: x = 0 (the y-axis). x-intercept: log2 x = 0 ⇒ x = 20 = 1, i.e. (1, 0).

(ii) log2 2 = 1 → (2, 1).   log2 4 = 2 → (4, 2).   log2 8 = 3 → (8, 3).   log2 (1/2) = −1 → (1/2, −1). Sketch: smooth curve through these and (1, 0), increasing slowly; approaches the y-axis from the right as x → 0+.

(iii) As x → 0+, log2 x → −∞ (the curve plunges down along the y-axis).   As x → ∞, log2 x → ∞, but very slowly (e.g. log2 106 ≈ 20).