Answer:
![4250 = 8500 e^{50 k}](https://tex.z-dn.net/?f=%204250%20%3D%208500%20e%5E%7B50%20k%7D)
And we can divide both sides by 8500 and we got:
![\frac{1}{2} = e^{50 k}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20%3D%20e%5E%7B50%20k%7D)
And we can solve for k using natural log
![k = \frac{ln(\frac{1}{2})}{50}= -0.0138629](https://tex.z-dn.net/?f=%20k%20%3D%20%5Cfrac%7Bln%28%5Cfrac%7B1%7D%7B2%7D%29%7D%7B50%7D%3D%20-0.0138629)
And we have the model give by:
![P(t) = 8500 e^{-0.01386294361 t}](https://tex.z-dn.net/?f=%20P%28t%29%20%3D%208500%20e%5E%7B-0.01386294361%20t%7D)
And if we replace t =97 we got:
![P(t=97) = 8500 e^{-0.01386294361 *97}= 2215.240](https://tex.z-dn.net/?f=%20P%28t%3D97%29%20%3D%208500%20e%5E%7B-0.01386294361%20%2A97%7D%3D%202215.240)
Step-by-step explanation:
For this case we can use the proportional model given by:
![\frac{dP}{dt}= kP](https://tex.z-dn.net/?f=%5Cfrac%7BdP%7D%7Bdt%7D%3D%20kP)
And we can rewrite the expression like this:
![\frac{dP}{P} = k dt](https://tex.z-dn.net/?f=%20%5Cfrac%7BdP%7D%7BP%7D%20%3D%20k%20dt)
If we integrate both sides we got:
![ln P = kt +C](https://tex.z-dn.net/?f=%20ln%20P%20%3D%20kt%20%2BC)
If we apply exponentials in both sides we got:
![P = e^{kt +C}](https://tex.z-dn.net/?f=%20P%20%3D%20e%5E%7Bkt%20%2BC%7D)
And we can rewrite the expression like this:
![P(t) = P_o e^{kt}](https://tex.z-dn.net/?f=%20P%28t%29%20%3D%20P_o%20e%5E%7Bkt%7D)
Where P represent the population and t the time in years since the starting value
For this case we have that ![P_o = 8500](https://tex.z-dn.net/?f=%20P_o%20%3D%208500)
And after t = 5*10 = 50 years the value is the half or 8500/2 = 4250
So we can use this condition and we have:
![4250 = 8500 e^{50 k}](https://tex.z-dn.net/?f=%204250%20%3D%208500%20e%5E%7B50%20k%7D)
And we can divide both sides by 8500 and we got:
![\frac{1}{2} = e^{50 k}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20%3D%20e%5E%7B50%20k%7D)
And we can solve for k using natural log
![k = \frac{ln(\frac{1}{2})}{50}= -0.0138629](https://tex.z-dn.net/?f=%20k%20%3D%20%5Cfrac%7Bln%28%5Cfrac%7B1%7D%7B2%7D%29%7D%7B50%7D%3D%20-0.0138629)
And we have the model given by:
![P(t) = 8500 e^{-0.01386294361 t}](https://tex.z-dn.net/?f=%20P%28t%29%20%3D%208500%20e%5E%7B-0.01386294361%20t%7D)
And if we replace t =97 we got:
![P(t=97) = 8500 e^{-0.01386294361 *97}= 2215.240](https://tex.z-dn.net/?f=%20P%28t%3D97%29%20%3D%208500%20e%5E%7B-0.01386294361%20%2A97%7D%3D%202215.240)