Answer:
![P(t) = 50000 e^(0.0069315 t)](https://img.qammunity.org/2021/formulas/mathematics/college/1hmirjb75xjl2gbezstcd69dprg78n2kvw.png)
Explanation:
For this case since the population follows an exponential model we have the general equation:
![P(t) = P_o e^(rt)](https://img.qammunity.org/2021/formulas/mathematics/college/ipvkdbvd8fqod1dry3c550rhpxwpjqsm6y.png)
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 :
and if we replace this into the general equation we have:
![50000 = P_o e^(r*0) = P_o](https://img.qammunity.org/2021/formulas/mathematics/college/78xombaheuaz920p5lju3siegbn1met0kq.png)
And the equation would be:
![P(t) = 50000 e^(rt)](https://img.qammunity.org/2021/formulas/mathematics/college/r7ecsv0edxm9pk9rlzdlkfpkhndqcyskcw.png)
Now we can use the second condition given
and replacing into the general formula we got:
![200000= 50000 e^(200t)](https://img.qammunity.org/2021/formulas/mathematics/college/vfygkfksndcrgby5810pb85d3ctqqxw27u.png)
We can divide both sides by 50000 and we got:
![4 = e^(200t)](https://img.qammunity.org/2021/formulas/mathematics/college/21un8ar1o27z0q1gp5q4v1vvcfain46pt9.png)
Now we can apply natural log on both sides:
![ln(4) = 200t](https://img.qammunity.org/2021/formulas/mathematics/college/b2lj9y0khtcj0do2r3vm0q80ywve7ut2zg.png)
And then:
![t = (ln(4))/(200)=0.0069315](https://img.qammunity.org/2021/formulas/mathematics/college/lw4hxnjebyovamjre6baghr2sro85cmn0f.png)
So then our final equation would be given by:
![P(t) = 50000 e^(0.0069315 t)](https://img.qammunity.org/2021/formulas/mathematics/college/1hmirjb75xjl2gbezstcd69dprg78n2kvw.png)