Don't know whether or not you've encountered differential equations yet, but will try that approach here.
The growth rate is dy/dt = ky (which states that the rate is proportional to the size of the population, y, and k is a constant.
Grouping like terms,
dy
--- = kt, so y = Ne^kt
y
We are told that at t=0, there are 880 bacteria. Thus, 880=N. Therefore,
y = 880e^(kt). After 5 hours the pop will be 4400; using this info, find k:
4400=880e^(5k), or 5 = e^(5k). So, our y = 880e^(kt) becomes
y = 880e^(5t).
What will be the pop after 2 hours? y(2)=880e^(10) = 880(22026) =
approx. 19,383,290 bacteria
Time to reach a pop of 2550? 2550 = 880e^(5t). Find t.
ln 2550 = ln 880 + 5t, so ln 2550 - ln 880 = 5t. Divide both sides by 5 to obtain this time, t.
