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orbie
 3 years ago
Problem attached below.
orbie
 3 years ago
Problem attached below.

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Outkast3r09
 3 years ago
Best ResponseYou've already chosen the best response.0if you look at all your answers...one sticks out of them. perhaps this will help \[log(a+b)=log(a)+log(b)\] \[log(a+b)=log(ba)=\frac{log(b)}{log(a)}\]

LolWolf
 3 years ago
Best ResponseYou've already chosen the best response.1The only solutions to the system are: \[ \log(a+b)=\log(a)+\log(b)\implies\\ b=\frac{a}{a1}, a\ne0, a1\ne0 \]

LolWolf
 3 years ago
Best ResponseYou've already chosen the best response.1@Outkast3r09 \( \log(a+b)=\log(a)+\log(b) \) Is *not* true for all cases.

orbie
 3 years ago
Best ResponseYou've already chosen the best response.0I understand the equations but I still can't find the right answers.

LolWolf
 3 years ago
Best ResponseYou've already chosen the best response.1The first and last are the only two.

orbie
 3 years ago
Best ResponseYou've already chosen the best response.0mmm...that is not right either according to my math hw program. This problem is such a hassle.

LolWolf
 3 years ago
Best ResponseYou've already chosen the best response.1Oh, I forgot to say that \(x>0\). The last two and the first are counterexamples to these.

LolWolf
 3 years ago
Best ResponseYou've already chosen the best response.1No, wait, never mind, it's only the first and the last.

LolWolf
 3 years ago
Best ResponseYou've already chosen the best response.1Because that restriction doesn't hold. I don't really know... you can check them all by hand, I guess, but I'm pretty sure it's only the first and last.

orbie
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you! It was the first and the last two that worked.

LolWolf
 3 years ago
Best ResponseYou've already chosen the best response.1Haha, yeah, I know why, now: the last one is imaginary, it *holds* but not in the real numbers.
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