Posted by Phillip on Sunday, March 28, 2010 at 6:55pm.
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.
Pre-Calc (exponential growth) - bobpursley, Sunday, March 28, 2010 at 7:14pm
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.
Pre-Calc (exponential growth) - Damon, Sunday, March 28, 2010 at 7:16pm
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
ln 10,000 = ln Po + 3 k
ln 40,000 = ln Po + 5 k
ln 10,000 -ln 40,000 = -2 k
ln (10,000/40,000) = -2k
-1.386 = -2k
k = .693
10,000 = Po e^(2.0794)
Po = 1250
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