Solving a Logistic Model - Population Data

[botty_12]

Hey

Im in the middle of modelling a logistic model off of population data but I am having a little bit of trouble. I am using a three parameter model
y=M/(1+Ce^-kt) and have set up three different equations to solve simultaneously. I originally used the value for when t=0 eliminating the k in one equation but I'm not too sure if i can use that to sub into the other equations. As far as solving it, I get to a point then become lost. Any help guys?

5.3= M/(1+Ce^-10k)

62.9=M/(1+Ce^-100k)

226.5=M/(1+Ce^-190k)

3.9 = M/(1+C)

 

[HallsofIvy]

Hey

Im in the middle of modelling a logistic model off of population data but I am having a little bit of trouble. I am using a three parameter model
y=M/(1+Ce^-kt) and have set up three different equations to solve simultaneously. I originally used the value for when t=0 eliminating the k in one equation but I'm not too sure if i can use that to sub into the other equations. As far as solving it, I get to a point then become lost. Any help guys?

5.3= M/(1+Ce^-10k)

62.9=M/(1+Ce^-100k)

226.5=M/(1+Ce^-190k)

3.9 = M/(1+C)[/QUOTE]

These are the equations you get when you substite t= 10, 100, 190, and 0, right? How did you get the y values on the left? You have 4 equations in 3 unknowns. If you are not absolutely certain that the function is of the given form, then you might not be able to find values of k, M and C that satisfy all 4. One thing you can do now is divide one equation by another, eliminating M.
Dividing the second equation by the first, 62.9/5.3= 11.87= (1+ Ce^-10k)/(1+ Ce^-100k) so 1+ Ce^-10k= 11.87+ 11.87Ce^-1o0k or
C(e^-10k- 11.87e^-100k)= 10.87.

Dividing the third equation by the first, 226.5/5.3= 42.75= (1+ Ce^-10k)/(1+ Ce^-190k) so 42.75- 42.75Ce^-190k= 1+ Ce^-10k or C(e^-10k- 42.75e^-190k)= 41.75.

Dividing one of those equations by the other eliminates C leaving a single equation in k.