# Two bacterial colonies

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• Nov 5th 2010, 02:15 PM
lancelot854
Would I multiply the entire problem by 2?
• Nov 5th 2010, 02:17 PM
Ackbeet
No, I don't think so. I would focus on the logarithm on the RHS. You can break that up according to the rules of logarithms.
• Nov 5th 2010, 02:22 PM
lancelot854
t/2 = log(base2)3 + log(base2)2^t/3
• Nov 5th 2010, 02:23 PM
Ackbeet
Simplify using more logarithm rules. What do you get?
• Nov 5th 2010, 02:25 PM
lancelot854
t/2 = log(base2)3 + t/3 log(base2)2
• Nov 5th 2010, 02:26 PM
Ackbeet
Keep going! You're 99.99% done.
• Nov 5th 2010, 02:29 PM
lancelot854
t/2 = log(base2)3 + t/3
• Nov 5th 2010, 02:29 PM
Ackbeet
So, t = ...
• Nov 5th 2010, 02:33 PM
lancelot854
For some reason I'm lost all of a sudden. :(
• Nov 5th 2010, 02:34 PM
Ackbeet
Bring the t/3 over to the LHS (to get all the t's on one side.)
• Nov 5th 2010, 02:37 PM
lancelot854
t/2 - t/3 = log(base2)3
• Nov 5th 2010, 02:38 PM
Ackbeet
Yep. Keep going!
• Nov 5th 2010, 02:42 PM
lancelot854
t/6 = log(base2)3
t =6 log(base2)3
• Nov 5th 2010, 02:49 PM
lancelot854
Yeah, finally it's solved. I concluded that the answer is approximately 9.51 hours. :)
I can't thank you enough for bearing with me and guiding me step by step. ^.^
• Nov 5th 2010, 03:05 PM
Ackbeet
You got it. I congratulate you on having the perseverance to keep going. You really did the majority of the work here, which is as it should be.

You're very welcome for my help. This is the kind of thing I love doing!
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