Asked by Oj

A colony of bacteria is grown under ideal conditions in a laboratory so that the population increases exponentially with time. At the end of 2 hours, there are 4,800 bacteria. At the end of 4 hours, there are 19,00 bacteria. How many bacteria were present initially?
Answered by Steve
A = P*e^kt

4800 = P*e^2k
19000 = p*e^4k

Now, 4800/p = e^2k, so
19000 = p*(4800/p)^2
19000 = 4800^2/p
p = 4800^2/19000 = 1212

just for grins, what's k?

4800=1212*e^2k
e^2k = 3.958
2k = ln 3.958 = 1.376
k = 0.688

so, A(x) = 1212*e^.688t
Answered by Anonymous
Bacteria experiment. If after one hour there were 1600 bacteria. Three hours later there was 400 bacteria. How many bacteria were there originally?
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