Given
The equation is given as
[tex]P(t)=1500e^{0.06t}[/tex]Explanation
To find the years required to exceed the population 1965.
[tex]\begin{gathered} 1965=1500e^{0.06t} \\ \frac{1965}{1500}=e^{0.06t} \\ 1.31=e^{0.06t} \end{gathered}[/tex]Take ln both sides.
[tex]\begin{gathered} ln1.31=0.06tlne \\ 0.27002=0.06t \\ t=\frac{0.27002}{0.06} \\ t=4.5 \end{gathered}[/tex]Hence the time required in years to exceed the population 1965 is
4.5 years.
