An epidemic follows the logistic growth model of I(t) = 5/(1+2e^(-kt)) k>0, wher
ID: 3347961 • Letter: A
Question
An epidemic follows the logistic growth model of I(t) = 5/(1+2e^(-kt)) k>0, where I(t) is the number of people in thousands who have been infected t weeks after the epidemic started.
a) Long term, what is happening to the number of people who have been infected?
b) If I(7)=2, what is k? Show Algebra and round to three decimals.
The answer for a is 5000 people will be infected and b is k=.041.
I am getting lost as to how to get to these two answers. Please show work in detail so I can follow. Thanks.
Explanation / Answer
a.) 5000
since t is long enough:
e^(-kt) =0
so I(t)= 5
and it is in thousands so its 5000
b.) 5/(1+2e^(-7k))=2
2.5= 1+2e^(-7k)
1.5/2= e^(-7k)
0.75= e^(-7k)
take log and get k= 0.041