Mathematics Advanced • Year 11 • Module 4 • Lesson 8
Change of Base & Logarithmic Equations
Apply logarithms to real exponential-growth contexts: compound interest, carbon dating, cooling, and a non-trivial double-base equation.
Problem 1 — How long to double an investment? (financial)
An amount P invested at 5% per annum, compounded annually, grows to A(n) = P × 1.05n after n years.
Set up: What are we solving for?
(i) Set A(n) = 2P and use logarithms to find the exact value of n (the doubling time). 2 marks
(ii) Use change of base or natural logarithms to evaluate n to 2 decimal places. 2 marks
(iii) Find the smallest whole number of years required for the investment to double, and explain why the answer to (ii) is non-integer but the practical answer must be an integer. 2 marks
Stuck? Take ln of both sides; 1.05n = 2 ⇒ n ln 1.05 = ln 2.Problem 2 — Carbon-14 dating (archaeology)
The fraction of original carbon-14 remaining in a sample after t years is
N(t)/N0 = (½)t / 5730, half-life = 5730 years
Set up: What are we solving for?
(i) An artefact contains 20% of its original C-14. Set up the equation N(t)/N0 = 0.20 and solve for t using logarithms. 3 marks
(ii) Use change of base or ln to give t to the nearest 10 years. 2 marks
(iii) A second artefact's measurement gives N/N0 = 1.1 (over 100% of original C-14 — impossible). Explain in one sentence what this means physically and what would have to happen to the laboratory reading for a real solution to exist. 2 marks
Problem 3 — Newton's law of cooling (forensic)
A cup of coffee with initial temperature 90°C in a 20°C room cools according to
T(t) = 20 + 70 e−kt, t in minutes, k a positive constant
After 5 minutes the coffee measures 70°C.
Set up: What are we solving for?
(i) Substitute T(5) = 70 to set up an equation in k, and solve for k using natural logarithms. Give k to 4 d.p. 3 marks
(ii) Using the value of k from (i), find the time required for the coffee to cool to 40°C. Round to the nearest minute. 3 marks
(iii) Explain in one sentence why the coffee can never actually reach the room temperature of 20°C, with reference to the structure of the model. 1 mark
Problem 4 — Bacterial growth (biology)
A bacterial colony doubles every 30 minutes. The population is modelled by
P(t) = P0 × 2t / 30, t in minutes
Set up: What are we solving for?
(i) Find the time required for the population to reach 1000 × P0. 3 marks
(ii) Convert your answer in (i) to hours and minutes (round to the nearest minute). 1 mark
(iii) An equivalent way to write the same model uses base e: P(t) = P0 × ekt. Find k (to 4 d.p.) by equating the two expressions. 3 marks
Stuck on (iii)? Use the fact 2t/30 = e(t/30) ln 2, so k = (ln 2)/30.Problem 5 — Equation with two different bases
An engineer needs to solve
3x + 1 = 52x
Set up: What are we solving for?
(i) Take natural logarithms of both sides and use the power law to bring down the exponents. 1 mark
(ii) Expand, collect the x-terms on one side, and solve for x in exact form (a single fraction of natural logs). 3 marks
(iii) Evaluate x to 3 decimal places and verify by substitution that both sides agree to 3 d.p. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Doubling time at 5% p.a.
Set up. We are using logarithms to solve an exponential equation for the time it takes a quantity to double.
(i) 1.05n = 2 ⇒ n ln 1.05 = ln 2 ⇒ n = ln 2 / ln 1.05.
(ii) n ≈ 0.6931 / 0.04879 ≈ 14.21 years.
(iii) After 14 full years the investment is at 1.0514P ≈ 1.98 P (not yet doubled); after 15 full years it is at 1.0515P ≈ 2.08 P. So the smallest whole-year answer is 15 years. The exact n ≈ 14.21 is non-integer because the model is continuous in n, but interest is credited only at year-end, so the practical answer must be rounded up.
Problem 2 — Carbon-14 dating
Set up. We use the change of base / natural log to solve a half-life equation for the elapsed time.
(i) (½)t/5730 = 0.20 ⇒ (t/5730) ln(½) = ln(0.20) ⇒ t = 5730 × ln(0.20) / ln(½) = 5730 × ln(0.20) / (−ln 2) = 5730 × ln 5 / ln 2.
(ii) t ≈ 5730 × 1.6094 / 0.6931 ≈ 5730 × 2.322 ≈ 13 300 years (to the nearest 10).
(iii) N/N0 = 1.1 means the sample now contains 10% more C-14 than at "time zero" — physically impossible because C-14 only decays (never grows) inside a dead sample. The reading must be a measurement or contamination error; the lab should re-measure or re-calibrate before any age can be reported.
Problem 3 — Newton's cooling
Set up. We are solving two exponential equations using natural logs to determine the cooling constant, then the cooling time.
(i) 70 = 20 + 70 e−5k ⇒ 50 = 70 e−5k ⇒ e−5k = 5/7 ⇒ −5k = ln(5/7) ⇒ k = −(1/5) ln(5/7) = (1/5) ln(7/5) ≈ (1/5)(0.3365) ≈ 0.0673 min−1.
(ii) 40 = 20 + 70 e−kt ⇒ 20 = 70 e−kt ⇒ e−kt = 2/7 ⇒ kt = ln(7/2) ⇒ t = ln(7/2) / k ≈ 1.2528 / 0.0673 ≈ 18.62, so t ≈ 19 minutes.
(iii) T(t) = 20 + 70 e−kt. As t → ∞, e−kt → 0 but never equals 0, so T → 20 from above but never reaches 20 — the room temperature is a horizontal asymptote of the model.
Problem 4 — Bacterial growth
Set up. We are solving exponential growth equations using logarithms, then converting between bases (2 and e).
(i) P0 × 2t/30 = 1000 P0 ⇒ 2t/30 = 1000 ⇒ (t/30) ln 2 = ln 1000 ⇒ t = 30 ln 1000 / ln 2 ≈ 30 × 6.908 / 0.693 ≈ 299.0 minutes.
(ii) 299 min = 4 h 59 min, i.e. 4 hours 59 minutes.
(iii) 2t/30 = e(t/30) ln 2, so the matching exponential model is P(t) = P0 × ekt with k = (ln 2)/30 ≈ 0.6931/30 ≈ 0.0231 min−1.
Problem 5 — Double-base equation
Set up. We are using ln of both sides (which works for any base) and the power law to convert an exponential equation into a linear one in x.
(i) ln(3x+1) = ln(52x) ⇒ (x + 1) ln 3 = 2x ln 5.
(ii) x ln 3 + ln 3 = 2x ln 5 ⇒ ln 3 = x(2 ln 5 − ln 3) ⇒ x = ln 3 / (2 ln 5 − ln 3) (equivalently x = ln 3 / ln(25/3)).
(iii) x ≈ 1.0986 / (3.2189 − 1.0986) = 1.0986 / 2.1203 ≈ 0.518. Verify: 31.518 ≈ 5.327 and 51.036 ≈ 5.327. ✓