Population Growth Question I am not sure how I am supposed to do this, but here is the problem and my attempt.

A farmer wants to produce chickens for the community. Let $P(t)$ represent the population of chickens at time $t$ in months. Assume the population of chickens reproduce at a rate proportional to its size. The farmer cultivates a small number of chickens for a trial run. In three months, the population of chickens increased from 75 to 300. The farmer wants to begin with an initial population of $P(0)$ chickens, and after a year, have $8000 + P(0)$ chickens, so that 8000 can be harvested, and $P(0)$ saved to start breeding again for the following year. What initial population $P(0)$ of chickens does the farmer need to start with?

Attempt:
First, since $P(t)$ represents the population at time $t$ in months, then the population growth equation can be written as: $$P(t) = P_0e^{kt}$$
Then I tried to find the relative growth rate. Since in three months, the population went from 75 to 300, then that means $$300 = 75e^{3k}$$
Or in other words, $k = \frac{\ln{4}}{3}$.
I am not exactly sure how to proceed from here, or if I am on the right track. I would like some assistance, or advice.
 A: Excellent so far!
You have $e^{3k}=4$ and you want $$Pe^{12k}=P+8000.$$
Note that $e^{12k}=4^4$.
Therefore  $256P=P+8000$ and $P=32$ are required.
A: Note that $k$ is the same in both scenarios because the chickens are the "same" type of chickens.
So, formulate $P(12) - P(0) = 8000 \implies P(0) \left( e^{12 k} - 1 \right) = 8000$, substituting $k = \frac{\ln 4}{3}$ and solve for $P(0)$.
A: Looks good so far!  This means that you have $$P(t) = P_{0}e^{\frac{t}{3}\ln 4}$$
or after rewriting:  $$P(t) = P_{0}e^{\ln 4^{t/3}} = P_{0}(4)^{t/3}.$$  Now it just remains to find the initial population $P_{0}$ so that $$8000 +P_{0} = P_{0}(4)^{12/3}.$$  Solving this equation for $P_{0}:$
\begin{align}
8000 +P_{0} &= P_{0}(4)^{12/3}\\
8000 &= P_{0}\left(4^{1/3} - 1\right)\\
P_{0} &=\frac{8000}{4^{12/3} - 1}.\\
\end{align}
Simplifying this gives us $$P_{0} = \frac{8000}{4^{4} - 1} = \frac{1600}{51} = 31.37...\approx 32.$$
A: While taking the logarithm is a good general strategy, in this case it is a needless complication. You have the population quadrupling in three months. So in a year, the population will quadruple four times ($12$ months divided by $3$ months is $4$). In other words, $P_{final}=4^4P_{initial}$. You also need $P_{final}=8000+P_{initial}$. Since both $4^4P_{initial}$ and $8000+P_{initial}$ are equal to $P_{final}$, they are equal to each other: $4^4P_{initial}=8000+P_{initial}$.
