Question

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.

Answer 1

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.

Answer 2

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