Exponential continuous growth $\mathrm{ln}a$ vs. $r$? Huh?

So, given a simple population continuous growth problem, it seems that the entirety of the internet uses $P={P}_{0}{e}^{rt}$ where $P$ is the population over time, ${P}_{0}$ is the initial population, $r$ is the percentage of change, and $t$ is time.

But this class is asserting that $P={P}_{0}{e}^{(\mathrm{ln}a)(t)}$ where $a=1+r$, and everything else stays the same.

So lets say a population starts at $10,000$ with $10$ continuous growth rate. What is the population in $10$ years?

The internet says that $P=10,000{e}^{(.10)(10)}=27,182$

But the class says that $P=10,000{e}^{(\mathrm{ln}1.10)10}=25,937$ ($a=1+r$ which is $1.10$)

but that equals the usual constant growth of $P={P}_{0}{a}^{t}=25937$

What the heck? What am I missing?

So, given a simple population continuous growth problem, it seems that the entirety of the internet uses $P={P}_{0}{e}^{rt}$ where $P$ is the population over time, ${P}_{0}$ is the initial population, $r$ is the percentage of change, and $t$ is time.

But this class is asserting that $P={P}_{0}{e}^{(\mathrm{ln}a)(t)}$ where $a=1+r$, and everything else stays the same.

So lets say a population starts at $10,000$ with $10$ continuous growth rate. What is the population in $10$ years?

The internet says that $P=10,000{e}^{(.10)(10)}=27,182$

But the class says that $P=10,000{e}^{(\mathrm{ln}1.10)10}=25,937$ ($a=1+r$ which is $1.10$)

but that equals the usual constant growth of $P={P}_{0}{a}^{t}=25937$

What the heck? What am I missing?