Answer:
[tex] P(t) = 50000 e^{0.0069315 t}[/tex]
Step-by-step explanation:
For this case since the population follows an exponential model we have the general equation:
[tex] P(t) = P_o e^{rt}[/tex]
Where P(t) represent the population at time t. t represent the years since the starting year.
r represent the growth/dcay constant rate
For this case we have the initial condition given : [tex] P(0) = 50000[/tex] and if we replace this into the general equation we have:
[tex] 50000 = P_o e^{r*0} = P_o[/tex]
And the equation would be:
[tex]P(t) = 50000 e^{rt}[/tex]
Now we can use the second condition given [tex] P(200) =200000[/tex] and replacing into the general formula we got:
[tex] 200000= 50000 e^{200t}[/tex]
We can divide both sides by 50000 and we got:
[tex] 4 = e^{200t}[/tex]
Now we can apply natural log on both sides:
[tex] ln(4) = 200t[/tex]
And then:
[tex] t = \frac{ln(4)}{200}=0.0069315[/tex]
So then our final equation would be given by:
[tex] P(t) = 50000 e^{0.0069315 t}[/tex]