SOLUTION: In 1995, 500 runners entered a marathon. In 1998, 20500 runners entered the race. Because of the limited number of facilities, the carrying capacity of the number of racers is 5850

Algebra ->  Graphs -> SOLUTION: In 1995, 500 runners entered a marathon. In 1998, 20500 runners entered the race. Because of the limited number of facilities, the carrying capacity of the number of racers is 5850      Log On


   

Question 980288: In 1995, 500 runners entered a marathon. In 1998, 20500 runners entered the race. Because of the limited number of facilities, the carrying capacity of the number of racers is 58500. The number of racers at any time, t, in years, can be modeled by the logistic function of p(t)=c/1+Be^kt
1. what is the value of c?
2. what is the value of b?
3. what is the value of k?
i know how to find the value of c and b but i need help on finding the value of K please, any help would be appreciated.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
You should find that c = 58500 and b = 116.


p%28t%29+=+c%2F%281%2BBe%5E%28k%2At%29%29


p%28t%29+=+58500%2F%281%2B116e%5E%28k%2At%29%29 Plug in c = 58500 and b = 116


20500+=+58500%2F%281%2B116e%5E%28k%2A3%29%29 Plug in t = 3 and p(t) = 20500 (since 3 years after 1995, the amount of runners is 20500)


20500%281%2B116e%5E%28k%2A3%29%29+=+58500


20500%2B2378000e%5E%283k%29+=+58500


2378000e%5E%283k%29+=+58500-20500


2378000e%5E%283k%29+=+38000


e%5E%283k%29+=+38000%2F2378000


e%5E%283k%29+=+0.01597981497057


3k+=+ln%280.01597981497057%29


3k+=+-4.1364289175233


k+=+-4.1364289175233%2F3


k+=+-1.37880963917443


k+=+-1.3788 (rounding to 4 decimal places)


So the function is p%28t%29+=+58500%2F%281%2B116e%5E%28-1.3788t%29%29