I can't find differences between $P(1+r)^n$ and $P(2.71828)^{rn}$ They told me $P(1 + r)^n$ can be used to calculate money interest, example:

You invert $15,000.00, 20% interest annual, for 3 years:
$15,000(1 + 0.20)^3$ = 25,920

And that $P(2.71828)^{rn}$ can be used to calculate population growth, example:

We have 500 bacterias, growing at the rate of 25% monthly, for 16 months:
$(500)(2.71828)^{(0.25)(16)}$ = roughly 26,968

I can't tell difference on when to use one over another. I mean different than seeing the keywords money and population. How do I tell when to use one over the another?
As far as I can tell, they both are something like:


*

*We have a starting value, a percent, a period of time


If something grows at a rate of 25% per month, then the amount after 16 months will be $
P\cdot1.25^{16}$.  This is true of money, of population, or of anything else.  The phrase "grows at a rate of 25% per month" means that at the end of every month the quantity is 25% bigger than it was at the beginning of the month.  The factor of increase for one month is therefore 1.25, and the factor of increase for 16 months is $1.25^{16}$.
It is always possible to rewrite the expression for quantity in a base other than base 1.25.  The usual choice is the natural base $e\approx2.718$.  In this base, the amount is given by $P\cdot e^{16k}$, where $k=\log_e1.25\approx0.2231$.  But $k$ is not the same thing as the rate.  Sometimes $k$ is referred to as the "rate under continuous compounding" or "continuous rate".  This terminology comes from the theory of compound interest.  So, for example, if you invest \$15000 at an annual rate of 20%, compounded continuously, then the amount after three years will be $15000\cdot e^{(3)(0.2)}$, whereas without the continuous compounding it will be $15000\cdot1.2^3$.  If you are not familiar with compound interest, you can simply think of $k$ as the base conversion factor - that is, the factor needed to convert from base 1.25 to base $e$.
You always need to be clear about what kind of rate you are dealing with.  If the word "rate" is used without a modifier such as "continuous", you should assume that it is the ordinary sort of rate described in the first paragraph above.
