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.