Mathematics Advanced • Year 11 • Module 4 • Lesson 15
Module Synthesis — Exponential and Logarithmic Functions
Practise HSC-style writing across the whole module, ending with a structured extended response that combines exponential modelling and calculus.
1. Short-answer questions
1.1 Solve e^(2x) − 7 e^x + 12 = 0 for x. Give exact answers. 3 marks Band 3-4
1.2 Differentiate y = ln( (x − 1)(x + 2) ) and simplify your answer as a single fraction. 3 marks Band 4
1.3 A bacterial culture grows according to P(t) = 1000 · e^(0.15 t).
(a) Find dP/dt at t = 4 hours and interpret in plain words.
(b) Find the doubling time. Round to the nearest 0.1 hour. 4 marks Band 4
2. Extended response
2.1 A protected colony of seabirds on an island has population P(t) (in hundreds) modelled by
P(t) = 12 · e^(0.04 t), where t is years since 2020.
Conservation managers want the long-run behaviour, the growth rate, the doubling time, and the maximum value of a derived "biodiversity score" S(t) = P(t) · e^(−0.06 t) that combines colony size with a slowly worsening external pressure.
(a) Predict P(t) at t = 20 (the year 2040), to the nearest bird. (Recall P is in hundreds.)
(b) Find dP/dt at t = 0 and at t = 20. Comment on whether the absolute growth rate is increasing or decreasing.
(c) Find the doubling time of P, to the nearest 0.1 year.
(d) Show that S(t) can be written as 12 · e^(−0.02 t). Hence find the rate of change of S at t = 10. Comment on its sign in one sentence.
(e) Show that S(t) has no interior maximum on t > 0, and state the maximum of S on the interval 0 ≤ t ≤ 30. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — P(20) = 12 · e^(0.8) ≈ 12 · 2.2255 ≈ 26.71 hundred = 2671 birds.
Part (b) — 1 mark
• 1 mark — dP/dt = 0.48 · e^(0.04 t). At t = 0: 0.48 (×100 birds/year = 48). At t = 20: 0.48 · e^(0.8) ≈ 1.068 (×100 = 106.8 birds/year). Notes "the absolute growth rate is increasing".
Part (c) — 1 mark
• 1 mark — t₂ = (ln 2) / 0.04 ≈ 17.33 ≈ 17.3 years.
Part (d) — 2 marks
• 1 mark — algebraically combines: S(t) = 12 · e^(0.04 t) · e^(−0.06 t) = 12 · e^(−0.02 t).
• 1 mark — dS/dt = −0.24 · e^(−0.02 t); at t = 10: dS/dt = −0.24 · e^(−0.2) ≈ −0.196 (hundred per year). The rate is negative, so the biodiversity score is decreasing in 2030.
Part (e) — 2 marks
• 1 mark — observes dS/dt = −0.24 e^(−0.02 t) < 0 for all t, so no interior stationary point ⇒ no interior maximum.
• 1 mark — S decreases throughout, so the max on [0, 30] is the left endpoint S(0) = 12 (i.e. 1200 birds in 2020).
Your response:
Stuck on (d)? Use a^m · a^n = a^(m+n) with a = e.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Solve e^(2x) − 7 e^x + 12 = 0 (3 marks)
Sample response. Let u = e^x. Then u² − 7u + 12 = 0 ⇒ (u − 3)(u − 4) = 0 ⇒ u = 3 or u = 4. Both are positive, so x = ln 3 or x = ln 4. x = ln 3 or x = ln 4 (≈ 1.099 or 1.386).
Marking notes. 1 mark — substitution u = e^x giving u² − 7u + 12. 1 mark — correct factorisation and both u-values. 1 mark — both x-values stated exactly with reverse-substitution. Numerical-only caps at 2/3.
1.2 — Differentiate y = ln((x − 1)(x + 2)) (3 marks)
Sample response. Split: y = ln(x − 1) + ln(x + 2). dy/dx = 1/(x − 1) + 1/(x + 2). Combine: dy/dx = [(x + 2) + (x − 1)] / [(x − 1)(x + 2)] = (2x + 1) / [(x − 1)(x + 2)]. (Valid for x > 1 so both factors are positive.)
Marking notes. 1 mark — splits using log laws before differentiating. 1 mark — correct sum of reciprocals. 1 mark — combined into single simplified fraction (2x + 1)/((x−1)(x+2)). Bonus context: domain x > 1 noted.
1.3 — Bacterial growth P(t) = 1000 e^(0.15 t) (4 marks)
(a) Sample response. dP/dt = 150 e^(0.15 t). At t = 4: dP/dt = 150 · e^(0.6) ≈ 150 · 1.822 ≈ 273 bacteria per hour. After 4 hours, the population is growing at about 273 bacteria per hour.
(b) Sample response. Doubling time t₂ = (ln 2) / 0.15 ≈ 0.6931 / 0.15 ≈ 4.62 ≈ 4.6 hours.
Marking notes. (a) 1 mark — correct derivative dP/dt = 150 e^(0.15 t) with chain factor 0.15 included. 1 mark — correct numerical answer ≈ 273 AND interpretive sentence. (b) 1 mark — formula (ln 2)/k. 1 mark — correct rounding to 4.6 h.
2.1 — Extended response (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). P(20) = 12 · e^(0.04 · 20) = 12 · e^(0.8) ≈ 12 · 2.2255 ≈ 26.706 hundred ≈ 2671 birds. [1 mark]
Part (b). dP/dt = 12 · 0.04 · e^(0.04 t) = 0.48 e^(0.04 t). At t = 0: dP/dt = 0.48 hundred/year = 48 birds/year. At t = 20: dP/dt = 0.48 · e^(0.8) ≈ 1.068 hundred/year ≈ 107 birds/year. The absolute growth rate is increasing, even though the percentage rate (k = 0.04) is constant — a hallmark of exponential growth. [1 mark]
Part (c). Doubling time: t₂ = (ln 2)/0.04 = 25 ln 2 ≈ 17.3 years. [1 mark]
Part (d). S(t) = P(t) · e^(−0.06 t) = 12 · e^(0.04 t) · e^(−0.06 t) = 12 · e^((0.04 − 0.06) t) = 12 · e^(−0.02 t). [1 mark — algebra]
dS/dt = 12 · (−0.02) e^(−0.02 t) = −0.24 e^(−0.02 t). At t = 10: dS/dt = −0.24 · e^(−0.2) ≈ −0.24 · 0.8187 ≈ −0.196 hundred per year (≈ −20 score-units per year). The sign is negative, so the biodiversity score is decreasing in 2030. [1 mark]
Part (e). dS/dt = −0.24 e^(−0.02 t). Since e^(−0.02 t) > 0 for all t, dS/dt < 0 for every t. So S has no interior stationary point on t > 0, and is strictly decreasing on [0, 30]. [1 mark] The maximum on [0, 30] therefore occurs at the left endpoint: S(0) = 12 (= 1200 birds × score-factor 1 in 2020). [1 mark]
Total: 7/7.
Band descriptors for marker.
Band 3: Computes P(20) and finds dP/dt; gets correct numbers but no interpretation; misses doubling time or doubling formula; ≈ 2-3 marks.
Band 4: All of (a), (b), (c) correct with clear working; (d) algebra mostly right but no comment on the sign of dS/dt; ≈ 4-5 marks.
Band 5: All five parts attempted; (e) attempts derivative argument but fails to conclude with endpoint reasoning; ≈ 5-6 marks.
Band 6: All parts complete, percent vs absolute growth distinction in (b), explicit "no interior stationary point" argument and endpoint max in (e). 7/7.