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
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.

- 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

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

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