Ayanna Goodman

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 $a\ast (1+k{)}^{t}$, 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?

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 $a\ast (1+k{)}^{t}$, 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?

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

$P(t)={P}_{0}\cdot {2}^{\frac{t}{4}}$

where ${P}_{0}$ is the initial population. In your non-e based formulation, this would be

$P(t)={P}_{0}\cdot ({2}^{\frac{1}{4}}{)}^{t}={P}_{0}\cdot (1+({2}^{\frac{1}{4}}-1){)}^{t}$

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:

$P(t)={P}_{0}\cdot {2}^{\frac{t}{4}}={P}_{o}\cdot {e}^{\mathrm{ln}2\cdot \frac{t}{4}}={P}_{0}\cdot {e}^{\frac{\mathrm{ln}2}{4}\cdot t}$

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 $\mathrm{ln}(\frac{1}{2})=-\mathrm{ln}2$

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

$P(t)={P}_{0}\cdot {2}^{\frac{t}{4}}$

where ${P}_{0}$ is the initial population. In your non-e based formulation, this would be

$P(t)={P}_{0}\cdot ({2}^{\frac{1}{4}}{)}^{t}={P}_{0}\cdot (1+({2}^{\frac{1}{4}}-1){)}^{t}$

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:

$P(t)={P}_{0}\cdot {2}^{\frac{t}{4}}={P}_{o}\cdot {e}^{\mathrm{ln}2\cdot \frac{t}{4}}={P}_{0}\cdot {e}^{\frac{\mathrm{ln}2}{4}\cdot t}$

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 $\mathrm{ln}(\frac{1}{2})=-\mathrm{ln}2$

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