#P(t)=Pe^(kt)............................(star)#
#"When "t=15, P(15)=500 rArr Pe^(15k)=500...........(1)#
#when t=40, P(40)=2000 rArr Pe^(40k)=2000.............(2)#
#(2)-:(1) rArr (Pe^(40k))/(Pe^(15k))=2000/500 rArr e^(25k)=4.#
#rArr (e^(25k))^(1/25)=4^(1/25) rArr e^k=4^(1/25)............(3)#
Now, rewriting #(1)" as, "P(e^k)^15=500", and now, using "(3),#
we get, #P(4^(1/25))^15=500 rArr P(4^(3/5))=500#
Hence, #P=500(4^(-4/5)).................(4)#
#(3),&, (4)", together with (star) "rArr P(t)=500(4^(-4/5))4^(t/25)#
Finally, to get the initial size of the culture, P(0) , we have to take
#t=0# in this last eqn., so that,
#P(0)=500(4^(-4/5))4^(0/25)=500(4^(-4/5))=((500)(4)(4^(-4/5)))/4#
#:. P(0)=(500/4)root(5)4=125root(5)4~~(125)(1.3195)=164.9375#
#:. P(0)~~165.#