Continuous growth and decay are modelled with N = Ae^(kt), where A is the initial value and k is a constant: positive for growth, negative for decay. To find the time for a value, isolate the exponential and take ln. Models of the form N = a ± be^(−kt) approach a limiting value a, and differentiating gives the rate of change, dN/dt = kN for the basic model.
Exponential models
- N = Ae^(kt): A is the initial value; k > 0 growth, k < 0 decay
- N = a ± be^(−kt) (k > 0): limiting value a as t → ∞
- Solving for t: isolate e^(kt), then take ln
- d/dt (Ae^(kt)) = kAe^(kt) = kN
The continuous model N = Ae^(kt)
A quantity growing or decaying continuously is modelled by N = Ae^(kt). When t = 0, e⁰ = 1, so A is the initial value. A positive k means growth; a negative k, or a term like e^(−kt) with k > 0, means decay. Substitute a value of t to find N at that time, keeping full calculator accuracy until the end.
Solving for the time
To find when a quantity reaches a particular value, first subtract any constant term, then divide to leave the exponential on its own, then take ln of both sides and divide by k. Do not take ln until the exponential is isolated; ln(a + b) cannot be split.
Limiting values in bounded models
Many models have the form N = a + be^(−kt) or N = a − be^(−kt), with k > 0. As t → ∞, e^(−kt) → 0, so N approaches a, the limiting value. The model starts at a ± b and gets closer and closer to a without reaching it, because be^(−kt) is always positive. Cooling liquids are a typical example: the temperature falls towards room temperature.
Finding k from data
The initial value gives A directly. A second data point then gives k: substitute it, isolate e^(nk), take ln and divide. Questions often ask you to show that k equals an exact expression such as (1/5) ln(9/4). If neither point is at t = 0, divide one equation by the other to eliminate A first.
Logistic models
Some populations follow a model like F = Ae^(kt)/(B + e^(kt)). At t = 0 it equals A/(B + 1). As t → ∞, dividing the top and bottom by e^(kt) shows that F → A, the limiting value. To find k from a data point, clear the fraction, collect the e^(kt) terms and take ln.
Rates of change
Differentiating brings the constant down: d/dt (Ae^(kt)) = kAe^(kt). So for the basic model, dN/dt = kN: the rate of change is proportional to the amount present, which is exactly what continuous growth means. Substitute a value of t to find the rate at that moment, and give units such as “per year”.
Worked examples
Example 1. A population is P = 5000e^(0.03t), where t is in years. State the initial population and find P after 20 years.
- At t = 0, P = 5000.
- At t = 20, P = 5000e^(0.6) = 9110.6…
Answer: Initially 5000; about 9110 after 20 years.
Marks: B1 for 5000, M1 for 5000e^(0.6), A1 for 9110.
Example 2. The temperature of a drink is θ = 25 + 50e^(−0.02t) °C after t minutes. Find t when θ = 40, to 3 s.f.
- Subtract 25: 50e^(−0.02t) = 15, so e^(−0.02t) = 0.3.
- Take ln: −0.02t = ln 0.3.
- t = ln 0.3 / (−0.02) = 60.198…
Answer: t = 60.2 minutes (3 s.f.).
Marks: M1 for e^(−0.02t) = 0.3, M1 for taking ln, A1 for 60.2.
Example 3. For N = 500 − 300e^(−0.1t), state the initial value and the limiting value, and explain why N never reaches the limit.
- At t = 0, N = 500 − 300 = 200.
- As t → ∞, e^(−0.1t) → 0, so N → 500.
- 300e^(−0.1t) is always positive, so N is always less than 500.
Answer: Initial value 200; limiting value 500.
Marks: B1 for 200, B1 for 500, B1 for the reason.
Example 4. N = Ae^(kt), with N = 40 when t = 0 and N = 90 when t = 5. Show that k = (1/5) ln(9/4).
- t = 0 gives A = 40.
- 90 = 40e^(5k), so e^(5k) = 9/4.
- Take ln: 5k = ln(9/4).
Answer: k = (1/5) ln(9/4), as required.
Marks: M1 for e^(5k) = 9/4, A1 for the result.
Example 5. For F = 400e^(kt)/(3 + e^(kt)), find the initial and limiting values, and, given F = 250 when t = 10, show that k = (1/10) ln 5.
- At t = 0: F = 400/4 = 100.
- As t → ∞, F → 400.
- Clear the fraction: 250(3 + e^(10k)) = 400e^(10k), so 750 = 150e^(10k).
- e^(10k) = 5, so 10k = ln 5.
Answer: Initial 100, limit 400, and k = (1/10) ln 5.
Marks: B1 for 100, B1 for 400, M1 for 750 = 150e^(10k), A1 for k.
Example 6. For P = 800e^(0.05t), find the rate of growth when t = 10, to 3 s.f.
- dP/dt = 800 × 0.05 × e^(0.05t) = 40e^(0.05t).
- At t = 10: 40e^(0.5) = 65.948…
Answer: 65.9 per year (3 s.f.).
Marks: M1 for dP/dt, A1 for 65.9.
Common mistakes
- Taking ln before isolating the exponential. Subtract and divide first, so that e^(kt) is on its own.
- Getting the sign of k wrong. Decay needs a negative exponent: e^(−kt) with k > 0, or a negative k.
- Saying the model reaches its limit. be^(−kt) is never zero, so the limiting value is approached but never reached.
- Rounding k before using it. Keep k exact, or to full calculator accuracy, until the final answer.
- Forgetting units for a rate. A rate of change has units, such as “per year” or “°C per minute”.
How exponential models are examined
Exponential modelling is a reliable source of Pure 3 (WMA13) marks in Edexcel International A-Level, and of Year 1 marks in Edexcel A-Level (9MA0). Questions usually ask you to:
- state an initial value and a limiting value, with a reason
- find a time for a given value, exactly or to a stated accuracy
- show that a constant k equals an exact log expression
- find a rate of change by differentiating
Modelling questions with full mark schemes are among the Pure 3 exponentials and logarithms exam questions. For more practice, use the exponentials and logarithms worksheet for Pure 3 or the exponentials and logarithms revision engine.
This uses eˣ and ln x throughout. The model N = Ae^(kt) also comes from solving dN/dt = kN in differential equations.
Frequently asked questions
What does A represent in N = Ae^(kt)?
The initial value: when t = 0, e⁰ = 1, so N = A.
How do you find the time in an exponential model?
Isolate the exponential term, take ln of both sides, then divide by the constant multiplying t.
What is a limiting value?
The value a model approaches as t becomes very large. For N = a ± be^(−kt), it is a.