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. ✓