There are many situations where the increase or decrease of some variable in a fixed time interval will be proportional to the magnitude of the variable at the beginning of that time interval.
For example, let’s look at a population of wee beasties which increases by 10% per year. If there were 100 wee beasties now, there would be an increase of 10 wee beasties after a year. We would see an increase of 500 wee beasties in a year when there were 5000 at the beginning.
Likewise, we can look at a population which decreases by 50% (i.e. a decrease to 1/2, or by a factor of 2) every day. A population of 100 would be down to 50 a day later, and a population of 5000 would drop to 2500 after one day.
These are all examples of exponential growth and decay:
A single equation can be used to solve all problems involving this type of change:
where ‘N’ is the number in the population after a time ‘t’, ‘‘ is the initial number, ‘k’ is the growth constant (if positive) or the decay constant (if negative), and ‘e’ is the base of natural logarithms (approximately 2.71828).
We can re-write this equation in another convenient form. Dividing the equation by Nº, and then taking the natural logarithms of both sides, we get
Note that in this form, we do not need to know the absolute values of ‘N’ or Nº’; all we need to know here is the ratio of these two values.