Question 980565: 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? 58500
2. what is the value of b? 116
3. what is the value of k? 1/3 ln 19/1189
4. in what year should at least 50500 runners be expected to run in the race?
i was able to solve for the above but i need help solving for the last part #4. thank you so much
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 1995
t=0
500=c/1+Be^kt
500=58500/(1+B), because e^kt=1 when t=0
500+500B=58500
500b=58500; B=116
20500=58500/{1+116*e^kt)}
0.35043=1/{1+116e^3k}
cross-multiply
1+116e^3k=2.8537
e^3k=1.8537/116=0.1598
ln of both sides
3k=-4.1364
k=-1.37881 (here I finally round).
50500=58500/{1+116e^3k}
cross-multiply
{1+116e^kt}=1.15842
subtract 1 and divide by 116
e^(-1.37881)t=0.001365;;ln of both sides
-1.37881t= -6.59612
divide by 1.37881 both sides
t= + 4.7839 years
By year 5 or (2000) at least 50,500 runners will be in the race. It would happen sooner if the race weren't run annually, but it will be at least by 2000.
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