Answer
$k$ $\approx$ $0.9605$ $years^{-1}$
$t$ $\approx$ $2.29$ years
Work Step by Step
$P'(t)$ = $kP(t)(\frac{1-P(t)}{20000})$
with general solution
$P(t)$ = $\frac{20000}{1-\frac{e^{-kt}}{C}}$
the initial condition $P(0)$ = $2000$ allow us to determine the value of C
$2000$ = $\frac{20000}{1-\frac{1}{C}}$
$C$ = $-\frac{1}{9}$
after one year, we know the population has grown to 4500 then
$4500$ = $\frac{20000}{1+9e^{-k}}$
$k$ = $\ln(\frac{81}{31})$ $\approx$ $0.9605$ $years^{-1}$
the population will increase to 10,000 at time t where P(t) = $10,000$ then
$10000$ = $\frac{20000}{1+9e^{-0.9605t}}$
$t$ $\approx$ $2.29$ years
