Answer
A population of 9000
Work Step by Step
The logistical differential equation is $\displaystyle P(t)=\frac{L}{1+Ce^{-kt}}$ where $P(t)$ is a population function of time, $L$ is the carrying capacity, and $C$ and $k$ are constants. To answer this question, we need to find the values of $L, C,$ and $k$. $L$ is actually given to us as 10,000 in this problem so now we only need to find $C$ and $k$. So far, our equation is:
$\displaystyle P(t)=\frac{10000}{1+Ce^{-kt}}$
At time $t = 0$ the population is 1000.
$\displaystyle P(0)=1000=\frac{10000}{1+Ce^{-(0)k}}$
$\displaystyle 1=\frac{10}{1+C(1)}$
$\displaystyle 1+C=10$
$C=9$
Now our equation is $\displaystyle P(t) = \frac{10000}{1+9e^{-kt}}$. At $t = 1$, the population is 2500.
$\displaystyle P(1) = 2500 = \frac{10000}{1+9e^{-k(1)}}$
$\displaystyle 1= \frac{4}{1+9e^{-k}}$
$\displaystyle 1+9e^{-k}= 4$
$\displaystyle 9e^{-k}= 3$
$\displaystyle e^{-k}= \frac{3}{9} = \frac{1}{3}$
$\displaystyle e^k = 3$
$k = \ln(3)$
Our equation is now: $\displaystyle P(t) = \frac{10000}{1+9e^{-\ln(3)t}} = \frac{10000}{1+\frac{9}{3^t}} = \frac{10000\cdot 3^t}{3^t+9}$. The question now asks what the population is 3 years after this(3 years after year 1), which is at $t = 4$.
$\displaystyle P(4) = \frac{10000}{1+9e^{-\ln(3)(4)}} = \frac{10000\cdot3^4}{3^4+9} = 9000$
