# Differential Equation, exponential growth

## Homework Statement

A colony of bacteria is grown under ideal conditions in a laboratory so that the population increases exponentially with time. At the end of 3 hours there are 10,000 bacteria. At the end of 5 hours there are 40,000. How many bacteria were present initially?

## The Attempt at a Solution

I have the following two equations setup:
10,000 = Ne^(3k)
40,000 = Ne^(5k)

I understand that this looks like a differential equation where I have to make them equal and solve for k (growth rate) and N (initial population) but I'm not quite sure how.

Matterwave
Gold Member
Well, if you divide the 2 equations, you should be able to eliminate N and get k. If you then plug in k to either equation, you should get N.

The differential equation part of the problem you have already solved by putting the equations in those forms.

Got it! Thank you so much.

40,000 = Ne^5k
10,000 = Ne^3k
_______________
4 = e^2k
ln4 = 2k
k = ln4/2

10,000 = Ne^(3*(ln4/2))
N = 1250

HallsofIvy
Actually, you could just "restart" the experiment at t= 3. That is write the equation as $N(t)= 10000e^{kt}$ where this "t" is now the number of hours beyond 3. Now you have only k to solve for: 5 hours after the start is 2 hours after 3 so, $40000= 10000e^{2k}$ and we have $e^{2k}= 4$, $2k= ln(4)$.