Respuesta :
Answer:
[tex] 4250 = 8500 e^{50 k}[/tex]
And we can divide both sides by 8500 and we got:
[tex] \frac{1}{2} = e^{50 k}[/tex]
And we can solve for k using natural log
[tex] k = \frac{ln(\frac{1}{2})}{50}= -0.0138629[/tex]
And we have the model give by:
[tex] P(t) = 8500 e^{-0.01386294361 t}[/tex]
And if we replace t =97 we got:
[tex] P(t=97) = 8500 e^{-0.01386294361 *97}= 2215.240[/tex]
Step-by-step explanation:
For this case we can use the proportional model given by:
[tex]\frac{dP}{dt}= kP[/tex]
And we can rewrite the expression like this:
[tex] \frac{dP}{P} = k dt[/tex]
If we integrate both sides we got:
[tex] ln P = kt +C[/tex]
If we apply exponentials in both sides we got:
[tex] P = e^{kt +C}[/tex]
And we can rewrite the expression like this:
[tex] P(t) = P_o e^{kt}[/tex]
Where P represent the population and t the time in years since the starting value
For this case we have that [tex] P_o = 8500[/tex]
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:
[tex] 4250 = 8500 e^{50 k}[/tex]
And we can divide both sides by 8500 and we got:
[tex] \frac{1}{2} = e^{50 k}[/tex]
And we can solve for k using natural log
[tex] k = \frac{ln(\frac{1}{2})}{50}= -0.0138629[/tex]
And we have the model given by:
[tex] P(t) = 8500 e^{-0.01386294361 t}[/tex]
And if we replace t =97 we got:
[tex] P(t=97) = 8500 e^{-0.01386294361 *97}= 2215.240[/tex]
