Question 980565
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.