Ayanna Goodman

## Answered question

2022-11-20

Why is e used in exponential growth functions?
Now I'm aware that ${e}^{x}$ is its own derivative, which makes it convenient to use in calculus. However, I have a question about this function:
Intuitively, an exponential growth function could be written as $a\ast \left(1+k{\right)}^{t}$, where a is the initial amount, k is the growth rate and t is the time.
However, it can also be rewritten as $a\ast {e}^{kt}$, same thing, but with Euler's number added to it.
In what way are these both the same? They yield the same answer. Why not 2, or 4, or 10? I know that one would need to change the growth constant if using one of those bases, the one I called k, but how come one doesn't need to change it when it's in base e?

### Answer & Explanation

yen1291kp6

Beginner2022-11-21Added 12 answers

The two k's in your formula will be different in order to make the two the same. Let's go with a simple concept in exponential growth:
Doubling time. As a concrete example, say we have a population that doubles every 4 units of time (Let's say days), so t is measured in days. Then it is pretty clear that the population at any given time is given by
$P\left(t\right)={P}_{0}\cdot {2}^{\frac{t}{4}}$
where ${P}_{0}$ is the initial population. In your non-e based formulation, this would be
$P\left(t\right)={P}_{0}\cdot \left({2}^{\frac{1}{4}}{\right)}^{t}={P}_{0}\cdot \left(1+\left({2}^{\frac{1}{4}}-1\right){\right)}^{t}$
thus making your growth rate k in this version ${2}^{\frac{1}{4}}-1$ Not very intuitive or easy to work with.
Changing it to base e just takes writing 2 as ${e}^{\mathrm{ln}2}$ and using exponent laws:
$P\left(t\right)={P}_{0}\cdot {2}^{\frac{t}{4}}={P}_{o}\cdot {e}^{\mathrm{ln}2\cdot \frac{t}{4}}={P}_{0}\cdot {e}^{\frac{\mathrm{ln}2}{4}\cdot t}$
This is the standard version of k we would use for this problem $k=\frac{\mathrm{ln}2}{4}$. So you see you can write it either way, you just need to change how you define k.
Using e is natural because we use these functions in differential equations all the time so using any other base would require constant change of bases. Finding k is easy, you just replace 4 with whatever the doubling time is in units of time. For exponential decay, you do the exact same but make it negative (because $\mathrm{ln}\left(\frac{1}{2}\right)=-\mathrm{ln}2$

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