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

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

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

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

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

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
(By my above solution)
 2 years ago

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

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

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

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Because 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.
 2 years ago

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

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