Mathematics Advanced • Year 11 • Module 4 • Lesson 1
Introduction to Exponential Functions
Apply y = ax to growth and decay scenarios: bacteria, depreciation, radioactive substances, and savings.
Problem 1 — Bacterial culture (growth)
A laboratory bacterial culture is modelled by
P(t) = 100 × 2t, t = hours since 9 am
Set up: What are we solving for?
(i) State the initial population at 9 am, and the value of the base. What does the base tell you about the population per hour? 2 marks
(ii) Predict the population at 2 pm (i.e. 5 hours later). Show one line of substitution and one of arithmetic. 2 marks
(iii) Explain in one sentence why the model is unrealistic for very large t, despite being a perfectly valid mathematical exponential function. 1 mark
Stuck? Revisit lesson § Worked Example 3.Problem 2 — Car depreciation (decay)
The retail value of a new car depreciates by 15% per year, modelled by
V(t) = 30 000 × (0.85)t, t = years from purchase
Set up: What are we solving for?
(i) Explain in one line why the multiplier is 0.85 (and not 0.15 or 1.15) for a 15% annual loss. 1 mark
(ii) Find V(2) and V(4) to the nearest dollar, and compute the dollar loss from year 2 to year 4. 3 marks
(iii) A buyer says: "Since the car loses 15% each year, after 7 years it will be worth zero." Without a calculation, explain why this is wrong (use the language of exponential decay). 2 marks
Problem 3 — Paper folding (recall extension)
A sheet of paper 0.1 mm thick is folded in half repeatedly. The thickness in millimetres after n folds is modelled by
T(n) = 0.1 × 2n, n = number of folds
Set up: What are we solving for?
(i) Calculate the thickness after 10 folds, in millimetres. 1 mark
(ii) Convert your answer to (i) into centimetres, then comment whether it matches the everyday intuition that "paper can't be folded more than 7 or 8 times". 2 marks
(iii) At 25 folds, the thickness is 0.1 × 225 mm. Express this in metres (use 210 ≈ 1000 to estimate), and explain in one sentence what this illustrates about exponential growth. 3 marks
Stuck? 225 = 210 × 210 × 25.Problem 4 — Radioactive substance (half-life)
A radioactive isotope halves every year. An initial sample of 800 g is described by
M(t) = 800 × (1/2)t, t = years since the measurement
Set up: What are we solving for?
(i) Identify the base, and state whether M(t) shows growth or decay. Justify in one phrase. 1 mark
(ii) Find the mass remaining after 1, 2 and 3 years. Write each as a fraction of 800 and as a number of grams. 3 marks
(iii) A second sample of 500 g of the same isotope is taken 1 year later than the first. Write a separate exponential model for the second sample (using t = years since the second sample was taken), and compare its "half-life" with the first. 2 marks
Problem 5 — Savings account (annual compounding)
$1000 is deposited in an account earning 6% per annum, compounded annually:
A(t) = 1000 × (1.06)t, t = years
Set up: What are we solving for?
(i) Compute A(1), A(5), A(10) to the nearest dollar. 3 marks
(ii) A student says: "After 10 years the interest is just 10 × 6% × 1000 = $600, so A(10) should be $1600." Explain in 1–2 lines why this reasoning is wrong and what the correct compounded answer (from (i)) shows. 2 marks
(iii) By trial, estimate the smallest whole number of years t for which A(t) exceeds $2000 (the deposit doubles). 2 marks
Stuck on (iii)? Calculate A(12), A(13)... until you cross $2000.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Bacterial culture
Set up. Compute the population over time and interpret the model's base.
(i) At t = 0 (9 am), P = 100 × 20 = 100 bacteria. Base = 2, meaning the population doubles each hour.
(ii) At 2 pm, t = 5: P(5) = 100 × 25 = 100 × 32 = 3200 bacteria.
(iii) In reality, bacteria run out of nutrients/space, so growth is not indefinite; the unconstrained exponential predicts an unrealistic population (e.g. at t = 30, P ≈ 1011).
Problem 2 — Car depreciation
Set up. Use a decay model with base 0.85 to estimate values at given times and reason about the long-run trend.
(i) Losing 15% leaves 85% = 0.85 of the previous year's value, so the multiplier per year is 0.85 (not 0.15, which would be the part lost).
(ii) V(2) = 30 000 × 0.7225 = $21 675. V(4) = 30 000 × (0.85)4 = 30 000 × 0.52200625 ≈ $15 660. Loss from year 2 → year 4: 21 675 − 15 660 = $6 015.
(iii) Exponential decay multiplies by the same factor (0.85) each year, so the value approaches 0 but never reaches it. After 7 years V(7) = 30 000 × (0.85)7 ≈ $9 605, still well above zero. The asymptote is y = 0; the car retains a fraction of value forever in the model.
Problem 3 — Paper folding
Set up. Apply T(n) = 0.1 × 2n for various fold counts and interpret.
(i) T(10) = 0.1 × 1024 = 102.4 mm.
(ii) 102.4 mm = 10.24 cm. Even after 10 folds the paper would be 10 cm thick — explaining the practical limit: the stack becomes physically too stiff to fold further.
(iii) 225 = 210 × 210 × 25 ≈ 1000 × 1000 × 32 = 32 000 000. So T(25) ≈ 0.1 × 3.2 × 107 mm = 3.2 × 106 mm = 3.2 km. After only 25 folds the stack would tower kilometres high — this is exponential growth: small initial value, modest exponent, astonishing output.
Problem 4 — Radioactive substance
Set up. Use the decay model M(t) = 800 × (1/2)t to compute remaining mass and to model a second sample.
(i) Base = 1/2. Since 0 < 1/2 < 1, M(t) shows decay.
(ii) M(1) = 800 × 1/2 = 400 g (1/2 of 800). M(2) = 800 × 1/4 = 200 g (1/4 of 800). M(3) = 800 × 1/8 = 100 g (1/8 of 800).
(iii) Second sample model: M2(t) = 500 × (1/2)t. The half-life is determined by the base (1/2), not the initial mass — both samples halve every year, even though they start at different amounts. Different initial mass, same half-life of 1 year.
Problem 5 — Savings account
Set up. Use A(t) = 1000 × (1.06)t to compute balances and reason about compounding.
(i) A(1) = 1000 × 1.06 = $1 060. A(5) = 1000 × 1.065 ≈ 1000 × 1.3382 ≈ $1 338. A(10) = 1000 × 1.0610 ≈ 1000 × 1.7908 ≈ $1 791.
(ii) The student computed simple interest (interest only on the original $1000). In compound interest, each year's interest is calculated on the new (larger) balance, so the interest itself grows year on year. The correct A(10) ≈ $1 791 exceeds the simple estimate of $1 600 by $191.
(iii) A(11) ≈ 1000 × 1.898 ≈ $1 898; A(12) ≈ $2 012. The deposit first exceeds $2 000 at t = 12 years.