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

## Answer this Question

## Related Questions

- Calculus - A colony of bacteria is grown under ideal conditions in a laboratory ...
- Calc - The number of bacteria in a culture is increasing according to the law of...
- Algebra - The number of bacteria in a certain population increases according to ...
- Pre-Calc/Trig... - is this the correct formula for me to solve this? A bacteria ...
- Calculus - The population of a colony of bacteria is modeled by the function p(...
- Pre Cal - A certain strain of bacteria divides every four hours. If a colony is ...
- Grade 12 Math-Urgent! - Bacteria have a doubling time of roughly 10 hours. A ...
- Math - Suppose that the number of bacteria in a certain population increases ...
- algebraic connections - At the start of an experiment, there are 100 bacteria. ...
- Math - Derivative of a polynomial function - The population, p in thousands of ...