How To Calculate Exponential Growth

Sometimes "exponential growth" is just a figure of speech, a reference to anything that grows unreasonably or unbelievably quickly. But in certain cases, we take the idea of exponential growth literally. For example, a population of rabbits can grow exponentially as each generation proliferates, then their offspring proliferate, and so on. Business or personal income can grow exponentially too through compound interest. When you're called upon to make real-world calculations of exponential growth, you'll work with three pieces of information: Starting value, rate of growth (or decay), and time.

To calculate exponential growth, use the formula

\(f(t) = a \times e^{kt}\)

where ‌a‌ is the initial value, ‌k‌ is a constant representing the rate of growth or decay (sometimes known as the growth factor), ‌t‌ is time, and ‌f‌(‌t‌) is the population's value at time ‌t‌. In this context, ‌e‌ is Euler's number, a mathematical constant that is very useful in exponential and logarithmic calculations.

The formula can also be represented as:

\(f(t) = a(1+r)^t\)

where ‌a‌ is the starting value, ‌r‌ is the rate of growth/decay, and ‌t‌ is the time elapsed (such as number of years or months).

Both of the formulas have some factor in the exponent that represents time, and they multiply some exponential operation with the rate of change by the initial amount to find the growth after a specified time period.

Imagine that a scientist is studying the growth of a new species of bacteria. While he could input the values of starting quantity, rate of growth and time into a population growth calculator, he decides to calculate the bacteria population's rate of growth manually using an exponential growth model.

Looking back on his meticulous records, the scientist sees that his starting population was 50 bacteria. Five hours later, he measured 550 bacteria.

Inputting the scientist's information into the equation for exponential growth or decay, he has:

\(f(t) = a \times e^{kt} \rightarrow 550 = 50 \times e ^{k5}\)

To begin solving for ‌k‌, first divide both sides of the equation by 50. This gives you:

\(\frac{550}{50} = 11 = e^{k5}\)

which simplifies to:

Next, take the natural logarithm of both sides, which is notated as ln(x). This gives you:

\(\ln{11} = \ln{e^{k5}}\)

The natural logarithm is the inverse function of ‌_e^x‌, so it effectively "undoes" the ‌e^x_‌ function on the right side of the equation, leaving you with:

\(\ln{11} = k5\)

Next, divide both sides by 5 to isolate the variable, which gives you:

\(k = \frac{\ln{11}}{5}\)

You now know the rate of exponential growth for this population of bacteria: ‌k‌ = ln(11)/5. If you're going to do further calculations with this population – for example, plugging the rate of growth into the equation and estimating the population size at ‌t‌ = 10 hours – it's best to leave the answer in this form. But if you're not performing further calculations, you can input that value into an exponential function calculator – or your scientific calculator – to get an estimated decimal value of 0.479579. Depending on the exact parameters of your experiment, you might round that to 0.48/hour for ease of calculation or notation.

Whether an initial population or quantity is growing or shrinking, there are countless examples of exponential growth in the world around us.

•Interest rates on loans is a very important, and often detrimental application of exponential growth as it builds upon previous interest values. It uses the exponential growth function to find the final value for simple interest – on a savings account for example.
•Half-life is an example of decaying exponential growth. As radioactive elements decay at a certain rate (this can be viewed as a nearly constant rate of decay), their ‌k‌ factor will determine how long it will be before only half of the original material remains.
•Doubling time applies to all types of exponential growth, and it is very similar to half-life. It describes the amount of time needed for exponential growth to result in the future value being double the amount of the starting value.