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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?

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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 $$P(t)=P(0)e^{(\ln(1.1))(t)}.$$

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  • $\begingroup$ so that is what the class is teaching. So where is the relationship between that and P=Po*e^(rt), because they don't calculate the same number? $\endgroup$ – user150153 May 15 '14 at 2:03
  • $\begingroup$ There are two common ways to measure rate of change. One of them is the instantaneous rate $r$ of change. That gives the formula $P(t)=P(0)e^{rt}$. Another is the rate of change over some interval of fixed length, in our case one year. That is common in specifying interest rates. Even if compounding is continuous, laws usually specify that the effective yearly interest rate be given to the customer. By generalizing the calculation I did in the answer, you can calculate the appropriate $\lambda$ is we are told that the city population increased by $p$ percent over $1$ year. $\endgroup$ – André Nicolas May 15 '14 at 2:19
  • $\begingroup$ (Cont.) You can generalize further by using the same idea to find the appropriate $\lambda$ if we are told that the population increased by $p$ percent over $k$ years. $\endgroup$ – André Nicolas May 15 '14 at 2:19
  • $\begingroup$ thanks Andre, I had thought that the issue was that one scenario was using r as a fixed change, and the other as an instantaneous change, and you confirmed it. two different r's. two different problems. Wish you could see this course. Last week it was "A car depreciates by $900 per year, so the value after t years is 900t + original value. (not negative 900 as it should have been). So I'm questioning this thing right and left. $\endgroup$ – user150153 May 15 '14 at 2:27
  • $\begingroup$ one question, using that method, computing annual and continuous growth, my continuous growth was less than the annual. Is that even possible? $\endgroup$ – user150153 May 15 '14 at 2:30

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