Very close, but not quite. Show me some steps.

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- Nov 5th 2010, 12:28 PMAckbeet
Very close, but not quite. Show me some steps.

- Nov 5th 2010, 12:35 PMlancelot854
Actually, all I did was replace the 1000 with 3000 and the 2 with 3...

- Nov 5th 2010, 12:38 PMAckbeet
Hmm. Turns out you can't do that. You know you've got the equation

$\displaystyle P_{2}(t)=3000(b^{t}).$

You can just plug in the 3000 for the 2000, because, as you can see, if you put in a 0 for the t, you just get $\displaystyle P_{2}(0)=3000$, which is correct for the second colony. You need to get another equation in order to solve for b. What are the conditions for the second colony? How can you get an equation from that condition? - Nov 5th 2010, 12:41 PMlancelot854
6000 = 3000b^3

2 = b^3

3rdsqrt(2) = b - Nov 5th 2010, 12:44 PMAckbeet
Excellent! In LaTeX, you can write it this way (double-click the equation to see how I wrote it):

$\displaystyle b=\sqrt[3]{2}.$

Another way to write it, which will actually be more useful to us, is this way:

$\displaystyle b=2^{1/3}.$

So now you've got

$\displaystyle P_{1}(t)=2000(2^{1/2})^{t},$ and

$\displaystyle P_{2}(t)=3000(2^{1/3})^{t}.$

What do you do now? - Nov 5th 2010, 12:47 PMlancelot854
2000(2^{1/2})^{t}=3000(2^{1/3})^{t}

- Nov 5th 2010, 12:47 PMAckbeet
Exactly. And turning the crank... Don't skip steps, now!

- Nov 5th 2010, 12:51 PMlancelot854
Now we just solve for t?

- Nov 5th 2010, 12:52 PMAckbeet
Yep. What do you get?

- Nov 5th 2010, 12:55 PMlancelot854
For some reason, the simplest problems always give me trouble. >:(

- Nov 5th 2010, 12:58 PMAckbeet
Oops. I made a mistake. It should be

$\displaystyle P_{1}(t)=1000(2^{1/2})^{t},$

$\displaystyle P_{2}(t)=3000(2^{1/3})^{t},$ leading to the equation

$\displaystyle 1000(2^{t/2})=3000(2^{t/3}).$

This is a nontrivial algebraic equation to solve. The first thing to do, I think, would be to divide both sides by 1000, giving you

$\displaystyle 2^{t/2}=3(2^{t/3}).$

Any ideas from here? - Nov 5th 2010, 01:01 PMlancelot854
My mind is blanking so much, I don't think we would divide by 3, for some reason I'm thinking log.

- Nov 5th 2010, 01:03 PMAckbeetQuote:

for some reason I'm thinking log.

- Nov 5th 2010, 01:05 PMlancelot854
log(base 2) 3(2^{t/3}) = t/2

- Nov 5th 2010, 01:10 PMAckbeet
A computer scientist, eh? For the computer scientist, log means base 2. For the scientist, it's base e. For the engineer, it's base 10. It doesn't really matter a whole lot which one you use, as long as your calculator can do it. I'd agree with your result, so long as you write the parentheses this way:

$\displaystyle \displaystyle\frac{t}{2}=\log_{2}\!\left(3(2^{t/3})\right).$

Now what can you do with this? (Hint: the logarithm of a product is the ... )