Answer :
We have an exponential growth problem here, it is given by the formula:
[tex]P(t)=P(0)\cdot e^{kt}[/tex]Where t is the time in years, P(0) is the initial population at t=0, and k is a constant.
We already know the population at t=0: P(0)=20,950.
And it doubles every 92 years, then when t=92, P(92)=2*P(0).
If we replace these values we can solve for k as follows:
[tex]\begin{gathered} P(0)=20,950\cdot e^{k\cdot0}=20,950 \\ P(92)=2\cdot20,950=41,900=20,950\cdot e^{k\cdot92} \\ We\text{ can solve for k as follows} \\ 41,900=20,950\cdot e^{k\cdot92} \\ \frac{41,900}{20,950}=e^{k\cdot92} \\ 2=e^{k\cdot92} \\ \text{Apply ln to both sides} \\ \ln 2=\ln e^{k\cdot92} \\ \text{Apply the properties of logarithms:} \\ 0.693=k\cdot92\cdot\ln e \\ \text{Simplify} \\ 0.693=k\cdot92 \\ k=\frac{0.693}{92} \\ k=0.0075 \end{gathered}[/tex]Then, to know the population 460 years from now, we need to replace k=0.0075 and t=460 and solve:
[tex]\begin{gathered} P(460)=20,950\cdot e^{0.0075\cdot460} \\ P(460)=20,950\cdot e^{3.466} \\ P(460)=20,950\cdot32 \\ P(460)=670,400 \end{gathered}[/tex]The population 460 years from now will be 670,400.
