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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 $a * (1+k)^t$, where a is the initial amount, k is the growth rate and t is the time.

However, it can also be rewritten as $a * e^{kt}$, 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 $e$?

Thank you.

Blauheim44
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    do you acknowledge that the $k$ in your first equation is different from the $k$ in your second equation? – okzoomer Dec 25 '21 at 09:21

1 Answers1

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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 $e^{\ln 2}$ and using exponent laws: $$P(t)=P_0\cdot 2^{\frac t 4}=P_o \cdot e^{\ln 2\cdot \frac t 4}=P_0\cdot e^{\frac {\ln 2} 4\cdot t}$$

This is the standard version of $k$ we would use for this problem $k=\frac {\ln 2} 4$. 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 $\ln (\frac 1 2)=-\ln 2$

Alan
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