Mathematics Advanced • Year 11 • Module 4 • Lesson 12
Differentiating loga x
Apply derivatives of general-base logarithms to scaling problems, rate-of-change and tangent questions in real contexts.
Problem 1 — Richter scale (base-10)
The Richter magnitude of an earthquake with seismograph amplitude A (in microns) is
M(A) = log10(A / A₀), for A > 0, A₀ a reference amplitude.
Set up: What are we solving for?
(i) Rewrite M(A) as a difference of two log10 terms, and hence find dM/dA. 2 marks
(ii) Evaluate dM/dA at A = 1000 microns. Give your answer to 3 significant figures. 2 marks
(iii) Use M(10A) − M(A) to show that a 10× increase in amplitude raises the magnitude by exactly 1 unit, no matter what amplitude you started at. 2 marks
Stuck on (iii)? log10(10A) − log10(A) = log10 10 = 1.Problem 2 — Information in bits (base-2)
The number of bits needed to encode N equally likely outcomes is
B(N) = log₂ N, for N > 0
Set up: What are we solving for?
(i) Find dB/dN as a simplified expression involving ln 2. 2 marks
(ii) Compute dB/dN at N = 64 (the size of a chess board's squares). Comment in one line on whether large or small N gives a steeper "cost per extra outcome". 2 marks
(iii) Show that B(2N) − B(N) = 1 exactly, for every N > 0. Interpret this result in plain English. 2 marks
Problem 3 — Tangent line in a log₃ chart
A scientist plots y = log₃ x for x > 0.
Set up: What are we solving for?
(i) Find the gradient of the tangent to y = log₃ x at x = 9. Leave ln 3 in the answer. 2 marks
(ii) Determine the y-coordinate at x = 9, then write the equation of the tangent in the form y = m(x − 9) + c. 3 marks
(iii) Sketch the relative position of the curve and the tangent at x = 9 in words: does the tangent lie above or below the curve for x ≠ 9? Justify in one sentence using the shape (concavity) of y = log₃ x. 2 marks
Stuck on (iii)? y = log₃ x is concave down for all x > 0.Problem 4 — pH revisited as a log₁₀ derivative
An equivalent definition of pH uses base 10 directly:
pH(H) = − log10 H, for H > 0
Set up: What are we solving for?
(i) Differentiate d(pH)/dH directly using d/dx(log10 x) = 1/(x ln 10). 2 marks
(ii) Evaluate d(pH)/dH at H = 10⁻⁷ (pure water). Give the answer in scientific form to 2 significant figures. 2 marks
(iii) Compare with d(pH)/dH at H = 10⁻⁴ (orange juice). State the ratio of the two derivatives and explain in one line what this tells you about how responsive pH is to changes in H. 3 marks
Problem 5 — Product of two logs
A composite scoring function used in education research is
S(x) = log₂ x · log₃ x, for x > 0
Set up: What are we solving for?
(i) Rewrite both factors using change of base into natural logs, and hence write S(x) as a constant times (ln x)². 2 marks
(ii) Use the chain rule to find S'(x), expressing your answer as a single fraction. 2 marks
(iii) Solve S'(x) = 0 and state the only stationary point in the domain. Use the sign of S'(x) on either side of this point to classify it. 3 marks
Stuck on (iii)? S'(x) involves a factor of ln x; set ln x = 0.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Richter scale
Set up. We differentiate magnitude with respect to amplitude and verify the "+1 per ×10" property.
(i) M(A) = log10 A − log10 A₀. Only the first term depends on A, so dM/dA = 1 / (A · ln 10).
(ii) At A = 1000: dM/dA = 1 / (1000 · 2.3026) ≈ 4.34 × 10⁻⁴ per micron.
(iii) M(10A) − M(A) = log10(10A/A₀) − log10(A/A₀) = log10(10A/A) = log10 10 = 1. Independent of the starting amplitude.
Problem 2 — Bits
Set up. We differentiate the bit-count and confirm the "+1 bit per doubling" property.
(i) dB/dN = 1 / (N · ln 2).
(ii) At N = 64: dB/dN = 1 / (64 · 0.6931) ≈ 0.0225 bits per extra outcome. The rate is much steeper at small N — adding an outcome to a 2-element set is far more "informative" than adding one to a 64-element set.
(iii) B(2N) − B(N) = log₂(2N) − log₂ N = log₂ 2 = 1. In plain English: doubling the number of equally likely outcomes always adds exactly one bit of information.
Problem 3 — Tangent to y = log₃ x at x = 9
Set up. We need a gradient, a point and a tangent equation, then a concavity statement.
(i) dy/dx = 1 / (x ln 3). At x = 9: dy/dx = 1 / (9 ln 3).
(ii) y = log₃ 9 = 2 (since 3² = 9). Tangent: y = (1 / (9 ln 3)) · (x − 9) + 2.
(iii) y = log₃ x is concave down on (0, ∞) (second derivative = −1/(x² ln 3) < 0). So the tangent at x = 9 lies above the curve for every x ≠ 9.
Problem 4 — pH from log₁₀
Set up. We differentiate the log10 definition of pH and compare two values.
(i) d(pH)/dH = − 1 / (H · ln 10) = −1 / (H ln 10).
(ii) At H = 10⁻⁷: d(pH)/dH = − 10⁷ / 2.3026 ≈ −4.3 × 10⁶ per mol/L.
(iii) At H = 10⁻⁴: d(pH)/dH ≈ −4.3 × 10³. Ratio = (10⁻⁷ derivative) / (10⁻⁴ derivative) = 10³. So at H = 10⁻⁷ the pH is 1000× more sensitive to changes in H than at H = 10⁻⁴ — small absolute changes in H near neutrality cause much larger pH swings than the same absolute changes near a more acidic solution.
Problem 5 — S(x) = log₂ x · log₃ x
Set up. We rewrite S using natural logs, differentiate, and classify the stationary point.
(i) S(x) = (ln x / ln 2)(ln x / ln 3) = (ln x)² / (ln 2 · ln 3). The constant factor is 1/(ln 2 · ln 3).
(ii) S'(x) = 2 ln x · (1/x) / (ln 2 · ln 3) = 2 ln x / [ x · ln 2 · ln 3 ].
(iii) ln 2 · ln 3 > 0 and x > 0, so S'(x) = 0 iff ln x = 0 iff x = 1. For 0 < x < 1, ln x < 0, so S'(x) < 0 (decreasing). For x > 1, ln x > 0, so S'(x) > 0 (increasing). Sign goes − then +, so x = 1 is a local minimum. (And S(1) = 0.)