 # 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=P0e^((lna)(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_0a^t=25937 What the heck? What am I missing? Ashlynn Hale 2022-08-15 Answered
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 $P={P}_{0}{e}^{\left(\mathrm{ln}a\right)\left(t\right)}$ 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 $P=10,000{e}^{\left(.10\right)\left(10\right)}=27,182$
But the class says that $P=10,000{e}^{\left(\mathrm{ln}1.10\right)10}=25,937$ ($a=1+r$ 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?
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Let us start from what everybody agrees on. The population $P\left(t\right)$ at time $t$ is given by
$\begin{array}{}\text{(1)}& P\left(t\right)=P\left(0\right){e}^{\lambda t},\end{array}$
where $\lambda$ is a constant.
Suppose that the population increases by $10\mathrm{%}$ in one year. Then putting $t=1$ in Equation (1), and using $P\left(1\right)=1.1P\left(0\right)$, we obtain
$1.1P\left(0\right)=P\left(0\right){e}^{\lambda }.$
From this we obtain that $1.1={e}^{\lambda }$. Taking natural logarithms, we find that $\lambda =\mathrm{ln}\left(1.1\right)$. Thus
$P\left(t\right)=P\left(0\right){e}^{\left(\mathrm{ln}\left(1.1\right)\right)\left(t\right)}.$