According to the exponential growth model, the population at any time instant is given by,
[tex]P(t)=P_{\circ}\cdot e^{kt}[/tex]
a)
Given that the initial population is 50,
[tex]\begin{gathered} P(0)=50 \\ P_{\circ}\cdot e^{k(0)}=50 \\ P_{\circ}\cdot e^{(0)}=50 \\ P_{\circ}(1)=50 \\ P_{\circ}=50 \end{gathered}[/tex]
Then the function becomes,
[tex]P(t)=50\cdot e^{kt}[/tex]
Given that after 7 years, there are 85 fish,
[tex]\begin{gathered} P(7)=85 \\ 50\cdot e^{k(7)}=85 \\ e^{7k}=1.7 \end{gathered}[/tex]
Taking logarithms on both sides,
[tex]\begin{gathered} \ln (e^{7k})=\ln (1.7) \\ 7k=\ln (7) \\ k=\frac{1}{7}\ln (7) \end{gathered}[/tex]
Substitute the value in the expression,
[tex]P(t)=50\cdot e^{\frac{1}{7}\ln (7)t}[/tex]
Thus, the required expression is,
[tex]P(t)=50\cdot e^{\frac{1}{7}\ln (7)t}[/tex]
If 'y' denotes the number of fish after time 't' years, then the expression becomes,
[tex]y=50\cdot e^{\frac{1}{7}\ln (7)t}[/tex]
b)
The population of fish after 12 years is calculated as,
[tex]\begin{gathered} P(12)=50\cdot e^{\frac{1}{7}\ln (7)12} \\ P(12)=50\cdot e^{\frac{1}{7}\ln (7)12} \\ P(12)\approx50\cdot e^{(3.335)} \\ P(12)\approx50\cdot(28.102) \\ P(12)\approx1405 \end{gathered}[/tex]
