Differential Equation, exponential growth

  • Thread starter dmitriylm
  • Start date
  • #1
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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?



Homework Equations





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.
 

Answers and Replies

  • #2
Matterwave
Science Advisor
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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.
 
  • #3
39
2
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
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
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Actually, you could just "restart" the experiment at t= 3. That is write the equation as [itex]N(t)= 10000e^{kt}[/itex] 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, [itex]40000= 10000e^{2k}[/itex] and we have [itex]e^{2k}= 4[/itex], [itex] 2k= ln(4)[/itex].

Now, you can answer "how many bacteria were present initially" by taking t= -3.
 

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