Given:
t = 11 years
[tex]P(t)=625\ln (2t+e)[/tex]
Asked: What will be the fish population in 11 years?
Solution:
In order to solve the fish population in 11 years, we will substitute 11 to t.
[tex]\begin{gathered} P(t)=625\ln (2t+e) \\ P(11)=625\ln (2\cdot11+e) \\ P(11)=625\ln (22+2.718281828) \\ P(11)=625\ln (24.718281828) \\ P(11)=625\cdot3.207543125 \\ P(11)=2004.714453 \end{gathered}[/tex]
ANSWER:
There will be 2005 fishes in 11 years.
