1. Excellent. So, if you use the template $P(t)=1000\,n^{rt},$ and you plug in for $P(t)$ and $t$, using the numbers you just gave me, what do you get?

2. $P(t) = 1000t$ or something like that, I'm never too sure.

3. Not quite. Compare these two equations with Post # 30:

$P(t)=1000\,n^{rt}$

$2000=1000\,n^{2r}.$ This equation is exactly the same equation as $P(2)=2000.$

Now how can you get $P(4)=4000$ to look like the second equation in this post?

4. $4000 = 1000n^4^r$

5. That's it! Now you have two equations and two unknowns. That is, you must solve

$2000=1000\,n^{2r},$

$4000=1000\,n^{4r}$ for $n$ and $r.$ How can you go about doing that?

6. Hang on. I've been leading you down a path that's more complicated than it needs to be. The reason is that

$n^{rt}=(n^{r})^{t}.$ I can define $b=n^{r},$ and I'm left with $b^{t}.$

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:

$P(t)=P_{0}\,b^{t}.$

You now have only one equation to solve:

$2000=1000\,b^{2}.$

7. First we would divide by 1000 to get $2 = n^2^r$. Couldn't we set this up into log form?
EDIT: Posted this before I read the previous post. Yeah I understand what you did there.

8. Check out Post # 36.

9. Yeah I understand post #36. Makes it much easier to follow.

10. So, what's b?

11. b = n^r

12. Right, but if you plug that into previous equations, you get the equation

$2000=1000\,b^{2}.$

That's one equation in one unknown. Solve for b. What do you get?

13. 2 = b^2
sqrt(2) = b

14. Excellent. So the first colony's population, which we'll now call $P_{1}(t),$ satisfies the equation

$P_{1}(t)=1000(\sqrt{2})^{t}.$

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?

15. P(t) = 3000(sqrt(3))^t

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