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Exponential growth and decay

Many real quantities grow or decay at a rate proportional to their size. They are modelled with e^kt, and you need to interpret the models.

The model N = Ae^kt

A is the initial value (at t = 0, e⁰ = 1).
k > 0 gives growth; k < 0 gives decay.
dN/dt = kAe^kt = kN: the rate of change is proportional to N.

To find k from data, substitute a known point and use ln.

The derivative of a^x

a^x = e^{x ln a}, so d/dx (a^x) = a^x ln a. For example, d/dx (2^x) = 2^x ln 2.

Interpreting

Initial value: put t = 0.
Long term: for T = 20 + 60e^-0.05t, e^-0.05t tends to 0, so T approaches 20 (room temperature).
Criticise a model: growth without limit is unrealistic; a refined model might include a limit.

Worked example

The temperature of a drink, T °C, t minutes after it is made, is T = 20 + 60e^-0.05t. Find the initial temperature and the time when T = 50.

t = 0: T = 20 + 60 = 80.
60e^-0.05t = 30, so e^-0.05t = 0.5.
-0.05t = ln 0.5, so t = 20 ln 2.

Answer: 80 °C; t = 13.9 minutes

Key idea: In N = Ae^kt, A is the start value and k the growth rate (negative for decay). Use ln to find t or k. d/dx (a^x) = a^x ln a.

Check yourself

A population is modelled by P = 300e^0.02t. Find P when t = 10, to 3 significant figures.
Show answer
366. P = 300e⁰.2 = 366.
The mass of a substance is m = 80e^-0.03t grams. Find how long it takes to halve, to 1 decimal place.
Show answer
23.1. 80e^-0.03t = 40 gives e^-0.03t = 0.5, so t = ln 2/0.03 = 23.1.
For P = 300e^0.02t, find the rate of increase dP/dt when t = 10, to 3 significant figures.
Show answer
7.33. dP/dt = 6e^0.02t, and at t = 10 this is 6e⁰.2 = 7.33.
Find the gradient of y = 2^x at x = 4, to 3 significant figures.
Show answer
11.1. dy/dx = 2^x ln 2. At x = 4: 16 ln 2 = 11.1.
T = 20 + 60e^-0.05t. What happens to T as t becomes very large?
- T approaches 80
- T approaches 0
- T keeps decreasing without limit
- T approaches 20
Show answer
T approaches 20. e^-0.05t tends to 0, so T tends to 20.
N = 50e^kt, and N = 80 when t = 5. Find k to 3 significant figures.
Show answer
0.0940. e^5k = 1.6, so 5k = ln 1.6 and k = 0.094.
A model predicts that a population grows as P = 300e^0.02t for all time. What is the main criticism?
- It predicts the population starts at zero
- It predicts the population decreases
- It cannot be used for t > 0
- It predicts the population grows without limit
Show answer
It predicts the population grows without limit. Food and space are limited, so a real population cannot grow for ever.
T = 20 + 60e^-0.05t. Find t when T = 30, to 1 decimal place.
Show answer
35.8. 60e^-0.05t = 10, so e^-0.05t = 1/6 and t = 20 ln 6 = 35.8.

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Last updated 7 November 2026. Spotted a mistake? Tell us.