# Two bacterial colonies

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• November 5th 2010, 01:28 PM
Ackbeet
Very close, but not quite. Show me some steps.
• November 5th 2010, 01:35 PM
lancelot854
Actually, all I did was replace the 1000 with 3000 and the 2 with 3...
• November 5th 2010, 01:38 PM
Ackbeet
Hmm. Turns out you can't do that. You know you've got the equation

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

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 $P_{2}(0)=3000$, 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?
• November 5th 2010, 01:41 PM
lancelot854
6000 = 3000b^3
2 = b^3
3rdsqrt(2) = b
• November 5th 2010, 01:44 PM
Ackbeet
Excellent! In LaTeX, you can write it this way (double-click the equation to see how I wrote it):

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

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

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

So now you've got

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

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

What do you do now?
• November 5th 2010, 01:47 PM
lancelot854
2000(2^{1/2})^{t}=3000(2^{1/3})^{t}
• November 5th 2010, 01:47 PM
Ackbeet
Exactly. And turning the crank... Don't skip steps, now!
• November 5th 2010, 01:51 PM
lancelot854
Now we just solve for t?
• November 5th 2010, 01:52 PM
Ackbeet
Yep. What do you get?
• November 5th 2010, 01:55 PM
lancelot854
For some reason, the simplest problems always give me trouble. >:(
• November 5th 2010, 01:58 PM
Ackbeet
Oops. I made a mistake. It should be

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

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

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

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

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

Any ideas from here?
• November 5th 2010, 02:01 PM
lancelot854
My mind is blanking so much, I don't think we would divide by 3, for some reason I'm thinking log.
• November 5th 2010, 02:03 PM
Ackbeet
Quote:

for some reason I'm thinking log.
A good reason. Try it!
• November 5th 2010, 02:05 PM
lancelot854
log(base 2) 3(2^{t/3}) = t/2
• November 5th 2010, 02:10 PM
Ackbeet
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:

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

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