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?