 # "How do Mathematicians determine when to use the constant e ? I'm only an undergrad so sorry if this is a dumb question, but I was studying Poisson distribution and it struck me that so many models involve ""e"". So it got me wondering; how/when/where/why do they decide to use it? I'm assuming they don't build a model and include ""e"" just because, and there must be some sort of fundamental intuition behind when its use is appropriate." blogswput 2022-09-14 Answered
How do Mathematicians determine when to use the constant e ?
I'm only an undergrad so sorry if this is a dumb question, but I was studying Poisson distribution and it struck me that so many models involve "e". So it got me wondering; how/when/where/why do they decide to use it? I'm assuming they don't build a model and include "e" just because, and there must be some sort of fundamental intuition behind when its use is appropriate.
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Actually, on the contrary - often you do use e "just because". Or, well, almost. The thing is, any exponential expression can be written as a power of e (for example ${2}^{3x-2}$ is just $\frac{1}{4}{e}^{3x\mathrm{ln}2}$). And powers of e are super convenient for all sorts of reasons - the first being that $\frac{d}{dx}{e}^{x}={e}^{x}$ and $\int {e}^{x}dx={e}^{x}+c$, so powers of e are ridiculously easy to differentiate and integrate. What that means is that whenever a mathematician is building a model that involves some sort of exponential, they'll usually convert it into a power of e unless they have a pressing reason not to.
EDIT: To walk through that example:
First, remember that ${a}^{b+c}={a}^{b}{a}^{c}$, and that for any a we have $a={e}^{\mathrm{ln}a}$ (this isn't a fancy property of e, just logarithms in general).
So
${2}^{3x-2}={2}^{3x}{2}^{-2}=\left({e}^{\mathrm{ln}2}{\right)}^{3x}{2}^{-2}$
Next, use the fact that $\left({a}^{b}{\right)}^{c}={a}^{bc}$, and that ${2}^{-2}=\frac{1}{4}$
$\left({e}^{\mathrm{ln}2}{\right)}^{3x}{2}^{-2}=\frac{1}{4}{e}^{\left(\mathrm{ln}2\right)\cdot 3x}$
Finally, rewriting it to be a little more readable, we get $\frac{1}{4}{e}^{3x\mathrm{ln}2}$