Answer
a. 7337 fish
b. about 1 year and 9 months.
Work Step by Step
(a)
$P(3)=\displaystyle \frac{10}{1+4e^{-0.8(3)}}=7.337$ (thousands)
After 3 years there are approximately 7337 fish .
(b) Solve for {\it t}.
$\displaystyle \frac{10}{1+4e^{-0.8t}}=5 \qquad $equate reciprocals
$\displaystyle \frac{1+4e^{-0.8t}}{10}=\frac{1}{5}\qquad .../\times 10$
$1+4e^{-0.8t}=2 \qquad .../-1$
$4e^{-0.8t}=1 \qquad .../\div 4$
$e^{-0.8t}=0.25 \qquad$ ... apply $\ln$() to both sides
$-0.8t=\ln 0.25 \qquad .../\div(-0.8)$
$t=\displaystyle \frac{\ln 0.25}{-0.8}\approx 1.733$
So the population will reach 5000 fish in about $1.73$ years
(0.73 years $\approx$ 9 months)