Posted by Phillip on .
A COLONY OF BACTERIA IS GROWN UNDER IDEAL CONDITIONS IN A LAB SO THAT THE POPULATION INCREASES EXPONENTIALLY WITH TIME. At the end of the three hours, there are 10,000 bacteria. At the end of the 5 hours, there are 40,000 bacteria. How many bacteria were present initially?
I was planning to use Y=P(O)e^kt
but I'm not sure how to set it up.

PreCalc (exponential growth) 
bobpursley,
Y(5)=40,000=Pe^5k
y(3)=10,000=Pe^3k
divide one equation by the other
4=e^2k solve for k. Put it back in, and solve for P(O)
There is an easier way. From the data, one can see that population doubles each hour. 
PreCalc (exponential growth) 
Damon,
Sure, use y = Po e^kt
when t = 3:
10,000 = Po e^3k
ln 10,000 = ln Po + 3k ln e but ln e is 1
so
ln 10,000 = ln Po + 3 k
similarly
ln 40,000 = ln Po + 5 k
subtract
ln 10,000 ln 40,000 = 2 k
ln (10,000/40,000) = 2k
1.386 = 2k
k = .693
then back
10,000 = Po e^(2.0794)
Po = 1250