# An application in growth

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• February 5th 2011, 04:49 PM
VonNemo19
An application in growth
The population of a bacterial culture grows at a rate proportional to the number of bacteria present at any time. After 3 hours it is observed that there are 400 bacteria present. After 10 hours there are 2000 present. What was the initial number of bacteria?

My attempt:

It is implied by context that a model for the rate of growth is

$\frac{dP}{dt}=kP$ which implies that $P(t)=ce^{kt}$.

But the fact that $P(3)=200$, and $P(10)=2000$ we see that

$400=ce^{3k}$ and $2000=ce^{10k}$

Solving for $c$ in the first equation gives

$c=400e^{-3k}$. Substituting into the second...

$2000=400e^{-3k}e^{10k}$ and we see that $k=\frac{\ln5}{7}$

Now, again using the fact that $P(3)=400$ we can conclude that $c=400e^{(-3\ln5)/7}$

So that $P(t)=400e^{\frac{\ln5}{7}(t-3)}$

and $P(0)=400e^{\frac{\ln5}{7}(0-3)}$

Have I done this right. If so, was there a faster way?
• February 5th 2011, 05:02 PM
Ackbeet
Everything looks good to me. The only thing that might have speeded things up a hair, would have been if you had divided one equation by the other. That would have gotten rid of the c a little quicker. There's not a lot of fluff in your solution, though.