So, given a simple population continuous growth problem, it seems that the entirety of the internet uses $P=P_0e^{rt}$ 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 $P=P_0e^{(\ln a)(t)}$ 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 $P=10,000e^{(.10)(10)}=27,182$
But the class says that $P=10,000e^{(\ln1.10)10} = 25,937$ ($a = 1 + r$ 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?
Let us start from what everybody agrees on. The population $P(t)$ at time $t$ is given by $$P(t)=P(0)e^{\lambda t},\tag{1}$$ where $\lambda$ is a constant.
Suppose that the population increases by $10\%$ in one year. Then putting $t=1$ in Equation (1), and using $P(1)=1.1P(0)$, we obtain $$1.1P(0)=P(0)e^{\lambda}.$$ From this we obtain that $1.1=e^{\lambda}$. Taking natural logarithms, we find that $\lambda=\ln(1.1)$. Thus $$P(t)=P(0)e^{(\ln(1.1))(t)}.$$