
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
$\displaystyle \frac{dP}{dt}=kP$ which implies that $\displaystyle P(t)=ce^{kt}$.
But the fact that $\displaystyle P(3)=200$, and $\displaystyle P(10)=2000$ we see that
$\displaystyle 400=ce^{3k}$ and $\displaystyle 2000=ce^{10k}$
Solving for $\displaystyle c$ in the first equation gives
$\displaystyle c=400e^{3k}$. Substituting into the second...
$\displaystyle 2000=400e^{3k}e^{10k}$ and we see that $\displaystyle k=\frac{\ln5}{7}$
Now, again using the fact that $\displaystyle P(3)=400$ we can conclude that $\displaystyle c=400e^{(3\ln5)/7}$
So that $\displaystyle P(t)=400e^{\frac{\ln5}{7}(t3)}$
and $\displaystyle P(0)=400e^{\frac{\ln5}{7}(03)}$
Have I done this right. If so, was there a faster way?

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.