Population changes with time Question


*

*The population of a certain community is known to increase at a rate proportional to the number of people present at time $t$. If the population has doubled in 5 years, how long will it take to triple? To quadruple?

*Suppose it is known that the population of the community in Problem 1 is 10,000 after 3 years. What was the initial population? What will be the population in 10 years?
Note - I need help with #2. I only included #1 because the first line of the second problem points to it.
The logistic equation demonstrated to us in class is 
$$
P =\frac{ak}{e^{-at}+bk}
$$
where $a$ = birth rate and $b$ = death rate. I assumed that the death rate is zero given the scope of this problem and attempted to solve for $a$, but I cannot isolate the variable. 
$$
-3a = ln(\frac{ak}{10000})
$$
Am I using the right equation? If not, then when is this one supposed to be used?
 A: Since the population increases at a rate proportional to the number of people present at time t (read: the rate of change of $P$ is proportional to $P$.)
$$\frac{\mathrm{d}P}{\mathrm{d}t} = kP$$
Separating the variables and integrating yields
$$\int \frac{1}{P} \, \mathrm{d}P = \int k \, \mathrm{d}t$$
So that we get
$$\ln P = kt + c$$ or equivalently, letting the initial population be $P_0$ so that when $t=0$, $P= P_0$ to find the arbitrary constant and re-arranging yields
$$P = P_0e^{kt}$$
We are given, that the population doubles ($2P_0$) in five years ($t=5$) so:
$$2P_0 = P_0 e^{5k}$$ 
Solving for $k$ yields
$$e^{5k} = 2 \implies k = \frac{\ln 2}{5}$$
Our population growth equation is then 
$$P = P_0 e^{\frac{\ln 2}{5}t}$$
So for the population to triple, we need to solve $3P_0 = P_0e^{kt}$ for $t$, where $k$ is what we found above. This yields 
$$e^{kt} = 3 \implies kt = \ln 3 \implies t = \frac{5\ln 3}{\ln 2}$$
The same can be done to find the time it takes for the population to quadruple, that is solve $4P_0 = P_0e^{kt}$ for t. 
For the second question, is we have that the population after three years ($t=3$) is $P = 10000$
then substituting this into our population growth equation gives 
$$10000 = P_0e^{3k}$$
so that we solve for $P_0$ to get
$$P_0 = \frac{10000}{e^{3k}} = \frac{10000}{\exp{\left(\frac{3\ln 2}{5}\right)}}$$
