Mathematics Advanced • Year 11 • Module 4 • Lesson 15
Module Synthesis — Exponential and Logarithmic Functions
Apply the full module's toolkit to mixed-context problems: investments, biology, mechanics, scaling and optimisation.
Problem 1 — When do investments overtake?
Investment A grows at 6% per annum compounded continuously: A(t) = 5000 · e^(0.06 t). Investment B grows annually at the same nominal 6%: B(t) = 5000 · (1.06)^t.
Set up: What are we solving for?
(i) Compute A(10) and B(10), rounded to the nearest dollar. Which is larger and by how much? 2 marks
(ii) Show that A(t) / B(t) = e^(0.06 t − t ln 1.06) and hence find the ratio at t = 30, to 3 significant figures. 3 marks
(iii) Explain in one sentence why A(t) > B(t) for every t > 0 (no calculation needed; use the inequality e^x > 1 + x for x > 0). 2 marks
Stuck? 0.06 > ln 1.06 because e^(0.06) > 1 + 0.06.Problem 2 — Two-dose drug schedule
A patient takes a 200 mg dose of a drug at t = 0. The amount remaining decays as D(t) = 200 · e^(−0.2 t) (t in hours). A second 200 mg dose is added to whatever is left at t = 6 hours.
Set up: What are we solving for?
(i) Find D(6) just before the second dose, to 1 decimal place. 1 mark
(ii) State the amount D(6⁺) immediately after the second dose, and write the formula for D(t) for t ≥ 6 (a fresh decay from this new starting value). 2 marks
(iii) Find the time after t = 6 at which D drops to 25 mg (a re-dose threshold). Give your answer to the nearest 0.5 hour. 3 marks
Problem 3 — Power-law from a log-log plot
A biologist plots metabolic rate R (in watts) against body mass M (in kg) on log-log paper for a range of mammals. The plotted points fall on a straight line with slope 0.75 and y-intercept (when log10 M = 0) equal to log10 3.4.
Set up: What are we solving for?
(i) Write the log-log relationship: log10 R = ____ + ____ log10 M. Then rewrite as R as a power of M. 2 marks
(ii) Use your model to predict R for M = 70 kg (roughly an adult human), to the nearest watt. 2 marks
(iii) Differentiate R = 3.4 M^(0.75) with respect to M, and interpret dR/dM at M = 70 in one sentence ("for each extra kg of body mass, the rate increases by ..."). 3 marks
Stuck? Use 10^(log10 a) = a to undo the log.Problem 4 — Where does e^x − a x bottom out?
For a fixed constant a > 0, consider f(x) = e^x − a x.
Set up: What are we solving for?
(i) Differentiate f and solve f'(x) = 0 to find the stationary x* in terms of a. 2 marks
(ii) Confirm it is a minimum using the second-derivative test. 2 marks
(iii) Find the minimum value of f in terms of a, simplifying as much as you can. For what value of a is the minimum value exactly 0? 3 marks
Problem 5 — Population reaches a cap
An island town's population (in thousands) is P(t) = 50 − 30 · e^(−0.1 t), where t is years since 2020. (As t → ∞, P → 50, the island's carrying capacity.)
Set up: What are we solving for?
(i) Find P(0) (the population in 2020) and state the limiting population as t → ∞. 2 marks
(ii) Differentiate P(t) to find dP/dt. Compute dP/dt at t = 10 and interpret in one sentence. 3 marks
(iii) Find when the population first reaches 45 thousand, to the nearest year. 2 marks
Stuck on (iii)? Solve 50 − 30 e^(−0.1 t) = 45 for t.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Investment comparison
Set up. We compare two compounding regimes at the same nominal rate.
(i) A(10) = 5000 · e^(0.6) ≈ 5000 · 1.8221 ≈ $9111. B(10) = 5000 · (1.06)^10 ≈ 5000 · 1.7908 ≈ $8954. A is larger by about $157.
(ii) A(t)/B(t) = (5000 e^(0.06 t)) / (5000 · (1.06)^t) = e^(0.06 t) / e^(t ln 1.06) = e^(0.06 t − t ln 1.06) = e^( (0.06 − ln 1.06) t ). With 0.06 − ln 1.06 ≈ 0.06 − 0.05827 = 0.001737. At t = 30: ratio = e^(30 · 0.001737) = e^(0.0521) ≈ 1.054.
(iii) Since e^x > 1 + x for every x > 0, we have e^(0.06) > 1.06, so e^(0.06 t) > (1.06)^t for all t > 0. Multiplying both sides by 5000 gives A(t) > B(t).
Problem 2 — Two-dose schedule
Set up. We compute the leftover before the second dose, then restart the decay.
(i) D(6) = 200 · e^(−1.2) ≈ 200 · 0.3012 ≈ 60.2 mg.
(ii) D(6⁺) = 60.2 + 200 = 260.2 mg. For t ≥ 6, D(t) = 260.2 · e^(−0.2 (t − 6)).
(iii) 25 = 260.2 · e^(−0.2 (t − 6)) ⇒ e^(−0.2 (t − 6)) = 25/260.2 ≈ 0.0961 ⇒ −0.2(t − 6) = ln 0.0961 ≈ −2.343 ⇒ t − 6 ≈ 11.72 ⇒ t ≈ 17.72 ≈ 17.5 hours after the initial dose (i.e. about 11.5 hours after the second dose).
Problem 3 — Log-log plot
Set up. We convert a straight-line log-log fit into a power law in R and M.
(i) log10 R = log10 3.4 + 0.75 · log10 M. Undo the log: R = 3.4 · M^(0.75).
(ii) R(70) = 3.4 · 70^(0.75). 70^(0.75) = e^(0.75 · ln 70) = e^(0.75 · 4.2485) = e^(3.186) ≈ 24.21. So R ≈ 3.4 · 24.21 ≈ 82 W.
(iii) dR/dM = 3.4 · 0.75 · M^(−0.25) = 2.55 · M^(−0.25). At M = 70: 2.55 / 70^(0.25) = 2.55 / 2.892 ≈ 0.88 W per kg. Each extra kilogram of body mass adds about 0.88 watts to the metabolic rate at this body size.
Problem 4 — Minimum of f(x) = e^x − a x
Set up. We optimise a one-parameter exponential function and find when the minimum is exactly zero.
(i) f'(x) = e^x − a. Set = 0 ⇒ e^x = a ⇒ x* = ln a.
(ii) f''(x) = e^x > 0 for all x, so x* is a minimum.
(iii) f(ln a) = e^(ln a) − a · ln a = a − a ln a = a(1 − ln a). Set this = 0: since a > 0, we need 1 − ln a = 0 ⇒ a = e. So the minimum value equals 0 exactly when a = e, and is positive for a < e, negative for a > e.
Problem 5 — Asymptotic population
Set up. We extract initial, limiting, growth-rate and threshold information from a "carrying-capacity" exponential.
(i) P(0) = 50 − 30 · 1 = 20 thousand. As t → ∞, e^(−0.1 t) → 0, so P → 50 thousand.
(ii) dP/dt = 30 · 0.1 · e^(−0.1 t) = 3 e^(−0.1 t). At t = 10: dP/dt = 3 e^(−1) ≈ 1.10 thousand per year. The town is gaining about 1100 people per year in 2030.
(iii) 45 = 50 − 30 e^(−0.1 t) ⇒ 30 e^(−0.1 t) = 5 ⇒ e^(−0.1 t) = 1/6 ⇒ −0.1 t = ln(1/6) = −ln 6 ⇒ t = 10 ln 6 ≈ 17.9 years ≈ 18 years (i.e. around year 2038).