Given the population growth formula as shown below
[tex]P=P_0e^{rt}[/tex][tex]\begin{gathered} r\rightarrow rate \\ P_0\rightarrow population\text{ at time t=0} \\ t\rightarrow time \\ \end{gathered}[/tex]When time is equal to zero
[tex]\begin{gathered} t=0 \\ P_0=100 \\ P=100 \end{gathered}[/tex][tex]\begin{gathered} when \\ t=3 \\ 800=100e^{r(3)} \\ 800=100e^{3r} \\ \frac{800}{100}=e^{3r} \end{gathered}[/tex][tex]\begin{gathered} e^{3r}=8 \\ lne^{3r}=ln8 \\ 3r=2.0794 \\ r=\frac{2.0794}{3}=0.6931 \end{gathered}[/tex]Thus
[tex]\begin{gathered} t=6 \\ P=100e^{0.6931\times6} \\ P=6398.19 \end{gathered}[/tex]Hence, the value of P(6) is approximately 6398.19
