Mathematics Advanced • Year 11 • Module 4 • Lesson 4
Exponential Modelling and Equations
Practise HSC-style writing on solving exponential equations and fitting Aekt models.
1. Short-answer questions
1.1 Solve 4x + 1 = 64. Show one line of matching-base working. 2 marks Band 3
1.2 A cup of coffee cools according to T = 85 e−0.1 t + 20 (t in minutes). Find the initial temperature and the temperature after 10 minutes. 3 marks Band 4
1.3 Solve 32x − 1 = 5 for x to 2 dp. State whether you used natural log or base-10 log and explain why your choice is valid. 4 marks Band 4
Stuck on 1.3? Either base works; take ln (or log), then isolate (2x − 1) by dividing by ln 3 (or log 3).2. Extended response
2.1 A pharmacist studies the elimination of a drug from the bloodstream. Two measurements are taken: at t = 0 the concentration is C(0) = 80 mg/L and at t = 6 hours it is C(6) = 20 mg/L. They assume the model C(t) = C0 e−kt (first-order elimination), with k > 0.
(a) Use the two measurements to find C0 and the elimination constant k. Give k to 3 dp.
(b) Find the half-life of the drug exactly (in terms of ln 2 and k) and to 1 dp in hours.
(c) The minimum effective concentration is 5 mg/L; below that, the drug stops working. Find, to 1 dp in hours, the latest time at which the drug is still effective. Express your answer both algebraically (using logs) and as a decimal.
(d) Explain in 1-2 sentences why a first-order model (constant k) is a reasonable starting point for many drugs, and identify one situation in which it would break down (in physiological terms). 7 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — identifies C0 = 80 from initial condition.
• 1 mark — substitutes (6, 20): 20 = 80 e−6k ⇒ e−6k = 0.25 ⇒ k = ln 4 / 6 ≈ 0.231.
Part (b) — 1 mark
• 1 mark — t½ = ln 2 / k ≈ 0.6931 / 0.231 ≈ 3.0 hours (a sensible answer, given C(6) = C(0)/4 = two half-lives).
Part (c) — 2 marks
• 1 mark — sets 5 = 80 e−k t, solves exactly: t = ln 16 / k.
• 1 mark — substitutes k ≈ 0.231: t ≈ 2.7726 / 0.231 ≈ 12.0 hours.
Part (d) — 2 marks
• 1 mark — explains that first-order kinetics (constant fractional elimination per unit time) often arises when the eliminating organ is far from saturation. 1 mark — identifies a specific failure case (e.g. saturable enzyme system, multiple-dose accumulation, alcohol's zero-order kinetics in liver) where k is not constant.
Your response:
Stuck on (c)? 5/80 = 1/16, so e−kt = 1/16 ⇒ −kt = −ln 16.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Solve 4x + 1 = 64 (2 marks)
Sample response. 64 = 43, so 4x + 1 = 43, giving x + 1 = 3, hence x = 2.
Marking notes. 1 mark — writes 64 as 43. 1 mark — equates exponents and solves. Students who jump to "take logs" reach the same answer (x = log4 64 = 3 − 1 = 2) but lose elegance; full marks if correct.
1.2 — T = 85 e−0.1 t + 20 (3 marks)
Sample response. Initial: T(0) = 85 · 1 + 20 = 105°C. T(10) = 85 e−1 + 20 ≈ 85 × 0.3679 + 20 ≈ 31.3 + 20 = 51.3°C.
Marking notes. 1 mark — substitutes t = 0 correctly to reach 105°C (common error: writes 85 + 0 + 20 = 105 from the wrong reason). 1 mark — substitutes t = 10. 1 mark — computes e−1 ≈ 0.368 and reaches ≈ 51.3°C.
1.3 — Solve 32x − 1 = 5 (4 marks)
Sample response. Take ln of both sides:
(2x − 1) ln 3 = ln 5 ⇒ 2x − 1 = ln 5 / ln 3 ≈ 1.6094 / 1.0986 ≈ 1.4650.
2x = 2.4650 ⇒ x ≈ 1.23.
I used natural log because the ex button on the calculator pairs naturally with the ln button — and either log will work because the ratio ln A / ln B equals log A / log B for the same A, B (the change-of-base rule).
Marking notes. 1 mark — takes log of both sides (any base). 1 mark — brings the exponent down using log(ab) = b log a. 1 mark — isolates x. 1 mark — correct decimal (≈ 1.23) and a sentence justifying log choice (any reasoning that mentions change-of-base or that "both natural and common logs give the same answer" earns this mark).
2.1 — Extended response (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a) — Fit C(t) = C0 e−kt.
At t = 0: C(0) = C0 · e0 = C0 = 80 mg/L. [1 mark.]
At t = 6: 20 = 80 e−6k ⇒ e−6k = 0.25 = 1/4 ⇒ −6k = ln(1/4) = −ln 4 ⇒ k = ln 4 / 6 ≈ 1.3863 / 6 ≈ 0.231 h−1. [1 mark.]
Part (b) — Half-life. ½ C0 = C0 e−k t½ ⇒ ½ = e−k t½ ⇒ k t½ = ln 2 ⇒ t½ = ln 2 / k ≈ 0.6931 / 0.231 ≈ 3.0 hours. (Sanity check: C(6) = C0/4 = two half-lives → 2 × 3 = 6 ✓.) [1 mark.]
Part (c) — Time at C = 5 mg/L. 5 = 80 e−kt ⇒ e−kt = 5/80 = 1/16 ⇒ −kt = ln(1/16) = −ln 16 ⇒ t = ln 16 / k. [1 mark — exact form.]
Numerically: ln 16 = 4 ln 2 ≈ 2.7726, so t ≈ 2.7726 / 0.231 ≈ 12.0 hours. (Confirms four half-lives: 80 → 40 → 20 → 10 → 5.) [1 mark.]
Part (d). First-order kinetics arises when the elimination pathway (typically liver enzymes or kidney filtration) is operating far below saturation, so the rate of removal is proportional to the current concentration — every doubling of concentration doubles the removal rate. [1 mark.] It breaks down when the eliminating system saturates (zero-order kinetics): a classic example is alcohol, eliminated by liver alcohol dehydrogenase, which saturates at modest blood-alcohol levels and then removes alcohol at a constant absolute rate (about 10 g/h) rather than at a constant fractional rate, so e−kt no longer fits. [1 mark — specific failure case identified.]
Total: 7/7.
Band descriptors for marker.
Band 3: Correct C0; attempts (b) but does not isolate k cleanly; little in (c)(d). ≈ 2-3 marks.
Band 4: Correct k and t½; correct (c) numerically but no exact form. (d) gives a vague reason. ≈ 4-5 marks.
Band 5: Full (a)(b)(c) including the exact form t = ln 16 / k. (d) explains first-order kinetics but does not name a saturation failure. ≈ 5-6 marks.
Band 6: All parts complete. (c) gives both exact (ln 16 / k) and numerical (12.0 h) forms. (d) names a specific physiological saturation case (e.g. alcohol, phenytoin, salicylate at high dose). 7/7.