Differential equations and exponential growth I was reading about differential equations and got stuck in a small detail that I can't make peace with.
If a population doubles every unit of time, I would write
$$ \frac{dP}{dt}=2P $$
which by separation of variables would yield
$$ P=P(0)e^{2t} $$
But I also know intuitively that I should have
$$ P = P(0)e^{(ln2)t} = P(0)2^{t}$$
which suggests the factor in the differential equation should be $\ln2$ instead of 2. What's wrong with my logic?
Thanks!
EDIT: This is the part of the textbook that confused me. 
Doesn't it confuse discrete and continuous cases as well?

 A: You are mixing discrete-time with continuous-time problems. For discrete-time problems, we use difference equations rather than differential equations. See https://en.wikipedia.org/wiki/Linear_difference_equation for more information on difference equations (or recurrence relations).
Your case corresponds to a geometric progression defined by the following recurrence relationship (or difference equation):
$$P_{t+1}=2P_{t} \implies P_{t}=P_02^t.$$
A: The differential equation:
$$\frac{dP}{dt}=2P$$
take into account infinitesimal growth of the population continuosly in time, whereas the equation:
$$P = P(0)2^{t}$$
is valid in a discrete time increment context.
A: Your second reasoning is correct. Exactly for the reason that you worked out. "The population doubles every unit of time" has the differential equation
$$
\frac{dP}{dt}=\ln(2)P
$$
More generally, "The population increases by r% every unit of time" has the continuous dynamical model
$$
\frac{dP}{dt}=\ln(1+r/100)P
$$
Or seen another way, doubling per time unit is equal to increase by a factor of $\sqrt 2$ every half time unit or by $2^{1/n}$ every $n$th part of a time unit. Which means that
$$
\frac{P((k+1)/n)-P(k/n)}{1/n}=n(2^{1/n}-1)P(k/n)
$$
and the limit of the factor in the last expression is $\ln(2)$.
A: The "intuitive" answer $P=P_02^n$ is only correct in a discrete system. It is the solution to the discrete functional equation $P_{n+1}=2P_n.$ If the population doubles at the end of every unit of time, then indeed the discrete solution is correct, where $n$ is the number of discrete time units. A natural number.
However in the differential equation $$\frac{dP}{dt}=2P$$ we are not saying that population doubles after a finite amount of time. Instead we're saying the instantaneous derivative matches twice the population. The proportional increase in population after an infinitesimal amount of time is an infinitesimal twice as big. $dP=2P\,dt.$ It's a totally different statement, and the intuition about discrete systems does not apply.
To relate a discrete time system to a continuous time system some limiting process has to take place, which is where the number $e$ comes in. If we imagine dividing up each time unit up into $n$ intervals, and let the population increase $2/n$-fold each interval, so after one sub interval the population goes to $P_0+2P_0/n$, etc, then at the end of the $n$ subintervals the population is 
$$
P=P_0\left(1+\frac{2}{n}\right)^n.
$$
The limit as $n\to\infty$ of the expression in the parentheses is $e^2$. That is essentially the definition of the number $e$. So what started as a doubling in the discrete case becomes and 7.38-fold increase in the continuum limit. And more generally, that's what the number $e$ does: it changes base discrete geometric growth $a^t$ into continuous exponential growth $e^{at}.$
Because of this translation between discrete and continuous, a continuous exponential growth problem which matches the discrete "doubling growth" problem at discrete times has to have $dP/dt=(\log2) P.$ 
Or alternatively, a discrete geometric growth problem that matches a continuous exponential growth problem with growth rate $2$ ($dP/dt=2P$) must follow $P_{n}=e^2P_{n-1}$ instead of doubling.
