Question

At any time t=>0 , in days, the rate of growth of a bacteria population is given by y' = ky, where y is the number of bacteria present and k is a constant.
The initial population is 1500 and the population quadrupled during the first two days. By what factor will the population have increased during the first three days?

Answer 1

quadrupled is doubled twice.
In two doubles, takes 4 days
doubles in 2 days.

What is 2^(3/2) ?

Answer 2

then y = c e^(kt)
when t = 4
4c = c e^(4k)
4 = e^ 4k
4k = ln4
k = ln4/4

y = 1500 e^((ln4/4)t)
so when t = 3
y = 1500 (e^(3ln4/4) = appr 4243

check:
at t=4
y = 1500 e^(4ln4/4) = 6000

Answer 3

Thanks!!