Exponential Growth and Decay
2026-02-28 17:42 Diff
https://brightchamps.com/en-us/math/numbers/exponential-growth-and-decay

The concept of exponential growth says that something always grows in relation to its current value. It is similar to the concept of doubling. For example, If a rabbit population doubles every month, the numbers would go 2, then 4, then 8, 16, 32, 64, 128, 256, and so on.


Let us consider a special tree that grows exponentially. Then, 


Its height in mm is ex. 


Where e stands for Euler’s number. 


E = 2.718


Therefore, let us now increase the value of x exponentially and see what happens. 

1-year-old \(e^1\) = 2.7 mm high, which is really tiny.

5-year-old \(e^5 \) = 148 mm high, which is as high as a cup. 

10-year-old \(e^{10}\) = 22 m high, which is as tall as a building.

15-year-old \( e^{15}\) = 3.3 km high, which is as tall as 10 stacked Eiffel towers.

20-year-old \(e^{20}\) = 485 km high, which reaches up into the space.


There is no tree that can grow that tall. Therefore, when someone says that it grows exponentially, we have to think about what they could be meaning. But sometimes, some things can grow exponentially, for a while. 


Therefore, we use a general formula, 


\(y(t) = a × e^{kt}\)

Where y(t) = value at time "t"

a = value at the start

k = rate of growth (when >0) or decay (when <0)

t = time


For example, 


Let us say that 2 months ago, we had 3 mice, and now we have 18. Find the value of the rate of growth. 

Let's start with the formula, 

\(y(t) = a × e^{kt}\)

We know that, 

a = 3, t = 2 and here, y(2) = 18.


Therefore, 

\(18 = 3 × e^{2k}\)

\(6 = e^{2k}\)

Taking the natural logarithm on both sides, 


ln(6) = ln(e2k)


ln(ex)=x, so:ln(6) = 2k


2k = ln(6)


\(k = \frac{ln(6)}{2}\)