Solution:
Given:
[tex]P(t)=600e^{0.06t}[/tex]
To get the time when the population will exceed 745:
[tex]\begin{gathered} when\text{ }P(t)=745, \\ 745=600e^{0.06t} \\ Divide\text{ both sides by 600;} \\ \frac{745}{600}=e^{0.06t} \\ 1.2417=e^{0.06t} \\ Take\text{ the natural logarithm of both sides;} \\ \ln1.2417=0.06t \\ Divide\text{ both sides by 0.06;} \\ \frac{\ln1.2417}{0.06}=t \\ t=3.608 \\ \\ To\text{ one decimal place;} \\ t\approx3.6 \end{gathered}[/tex]
Therefore, the time it will take for the population to exceed 745 is 3.6
