Problem: Population growth is proportional to the size of the population. From time $t=0$ to $t=1$, the population increases two times. At $t=2$, the population is $10000$. Find the population at $t=0$.
My take on it:
$P'=kP$
$P'-kP=0 \ \ | \cdot e^{-kt}$
$(Pe^{-kt})'=0$
$Pe^{-kt}=C $
$P(t)=C\cdot e^{kt}$
$P(0)=C=a $ (given notation)
$P(1)=Ce^{kt}=2a$
$C=2a\cdot e^{-k}$
$P(2)=10000=2a\cdot e^{-k}\cdot e^{2k}$
$10000=2a \cdot e^k$
Take natural logarithm on both sides,
$\ln10000=\ln(4a)\cdot k$
So this is where I have problem, I guess it's because of my lack of knowledge about logarithms. I assumed that I can take $2a$ out of $\ln$ sign as it is a constant, but then the solution is not right.
How should I do it?
The solution
$$ P(t) = Ce^{kt}$$
At time 0 and 1,
$$P(0) = C, \>\>\> P(1) = C e^{k}$$
Then, take the ratio $$ \frac{P(1)}{P(0)}= e^k=2$$
At time 2,
$$ P(2) = Ce^{2k} = P(0)e^{2k}= 10000$$
Thus, the population at time 0 is, $$P(0) =\frac{10000}{e^{2k}}= \frac{10000}{2^2}=2500$$