Excellent. So, if you use the template $\displaystyle P(t)=1000\,n^{rt},$ and you plug in for $\displaystyle P(t)$ and $\displaystyle t$, using the numbers you just gave me, what do you get?
Not quite. Compare these two equations with Post # 30:
$\displaystyle P(t)=1000\,n^{rt}$
$\displaystyle 2000=1000\,n^{2r}.$ This equation is exactly the same equation as $\displaystyle P(2)=2000.$
Now how can you get $\displaystyle P(4)=4000$ to look like the second equation in this post?
Hang on. I've been leading you down a path that's more complicated than it needs to be. The reason is that
$\displaystyle n^{rt}=(n^{r})^{t}.$ I can define $\displaystyle b=n^{r},$ and I'm left with $\displaystyle b^{t}.$
Thus, mathematically, we don't gain any flexibility in constructing our function by having both n and r available.
Bottom line: change our initial function to this:
$\displaystyle P(t)=P_{0}\,b^{t}.$
You now have only one equation to solve:
$\displaystyle 2000=1000\,b^{2}.$
Did you follow all that?
Excellent. So the first colony's population, which we'll now call $\displaystyle P_{1}(t),$ satisfies the equation
$\displaystyle P_{1}(t)=1000(\sqrt{2})^{t}.$
In order to solve the problem in the OP, you're going to need to repeat this process of finding the equation for the second colony's population. What do you get for that?