Most of these type problems use the exponential function P = Poekt
where p = population at time t
Po = the population at time 0
k = the time constant
t = time
Using the initial conditions, 2Po = Poekt then 2 = ek(24) then k = ln(2)/24 = .02888
So, in 2 years P(2) = (16,237)e(.02888)(2) = 17,202.5