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?

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- Nov 5th 2010, 11:40 AMAckbeet
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?

- Nov 5th 2010, 11:43 AMlancelot854
$\displaystyle P(t) = 1000t$ or something like that, I'm never too sure.

- Nov 5th 2010, 11:48 AMAckbeet
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? - Nov 5th 2010, 11:52 AMlancelot854
$\displaystyle 4000 = 1000n^4^r$

- Nov 5th 2010, 11:54 AMAckbeet
That's it! Now you have two equations and two unknowns. That is, you must solve

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

$\displaystyle 4000=1000\,n^{4r}$ for $\displaystyle n$ and $\displaystyle r.$ How can you go about doing that? - Nov 5th 2010, 12:02 PMAckbeet
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? - Nov 5th 2010, 12:02 PMlancelot854
First we would divide by 1000 to get $\displaystyle 2 = n^2^r$. 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. - Nov 5th 2010, 12:03 PMAckbeet
Check out Post # 36.

- Nov 5th 2010, 12:07 PMlancelot854
Yeah I understand post #36. Makes it much easier to follow.

- Nov 5th 2010, 12:08 PMAckbeet
So, what's b?

- Nov 5th 2010, 12:12 PMlancelot854
b = n^r

- Nov 5th 2010, 12:14 PMAckbeet
Right, but if you plug that into previous equations, you get the equation

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

That's one equation in one unknown. Solve for b. What do you get? - Nov 5th 2010, 12:18 PMlancelot854
2 = b^2

sqrt(2) = b - Nov 5th 2010, 12:20 PMAckbeet
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? - Nov 5th 2010, 12:27 PMlancelot854
P(t) = 3000(sqrt(3))^t