Exponential Growth & Decay

Exponential growth and decay is where a function’s growth or decay rate is proportional to the function’s current value. Firstly, we can see exponential growth in things like an increasing population. That is, population size grows at a rate proportional to the number currently in the population. Secondly, exponential decay occurs when the decay rate is proportional to the function’s current value. For example, we can see this in radioactive decay or drug concentration in a bloodstream.

In a continuous setting,  we can use e^x to model exponential growth, whereas we should use e^{-x} to model exponential decay. Note that mathematical modelling has drawbacks. That is, models often need refinements and fail when populations become unpredictable. See the problems and pitfalls of mathematically modelling the 2020 Coronavirus spread.

See Example 1 for more on Exponential Growth and Example 2 for Exponential Decay.

Geometric Growth & Decay

Geometric growth and decay is the same as exponential growth and decay except the function is only evaluated at discrete values. For example, the geometric series with a start value of 5 and a common ratio of 2, i.e. 5+10+20+40+80+... is an example of a series that exhibits exponential growth discretely. This is geometric growth. However, the series 10+1+0.1+0.01+0.001+... is an example of a series that exhibits exponential decay discretely. This is geometric decay.

Compound interest is an example of geometric growth that you have seen before. Compound interest is where we earn interest on interest. That is, if a savings account has a an interest rate of 5% per annum and £2000 is deposited into the savings account, after one year there will be 2000\times 1.05=2100. Then there will be 2000\times 1.05^2=2205 after two years. Generalising, there will be 2000\times 1.05^n after n years.

At this level, you will mostly study exponential growth and decay.


The number of cases N for a given virus t days after the onset of an outbreak in a given country can be modelled using the equation N(t)=N_0e^{0.25t}.

  1. Given that the number of people initially diagnosed with the virus is 4, what is the value of N_0?
  2. How many people are likely to have the virus on day 10?
  3. Sketch the curve of N against t.
  4. Explain why the curve is unlikely to continue to follow this model.
  5. Another virus has a faster rate of spread. Suggest how you could refine the model above to show this.


  1. Substituting t=0 and setting N(0)=4, we have that N_0=4.
  2. N(10)=48.73 to two decimal places. The number of people with the virus on day 10 is about 49.
  3. growth and decay
  4. The government in this country will take action to ‘flatten the curve’.
  5. Increase the value from 0.25 in the exponent.

A rare species of dandelion is being observed. The population, p, after a given number of years, t, is given by p(t)=500e^{-0.3t}.

  1. Find the initial population of dandelion.
  2. Find the rate at which the dandelion population changes after 5 years. Give your answer to two decimal places.
  3. Explain why the population of dandelion, provided that it follows this model, will eventually die out.


  1. Initial population is the number of dandelion at the start of the study, that is, when t=0. Hence, it can be found by evaluating p(0)=500.
  2. We can find the rate at which the population changes after 5 years by substituting t=5 into the expression for \frac{dp}{dt} where \frac{dp}{dt}=-0.3\times 500e^{-0.3t}=-150e^{-0.3t}. Substituting t=5 into this expression gives \frac{dp(5)}{dt}=-150e^{-0.3\times 5}=-33.47. This means that the dandelion are dying at a rate of 33.47 per year after 5 years. See more on differentiating e to the x.
  3. As t becomes larger e^{-t} becomes very small eventually approaching zero.



Using simultaneous equations to find (and then interpreting) the unknown constants that determines the exponential growth in the value of a vintage car.