TITLE: Differential equations and exponential growth
QUESTION [5 upvotes]: 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?

REPLY [4 votes]: 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 $\ln(2)$.