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Outkast3r09Best ResponseYou've already chosen the best response.0
if you look at all your answers...one sticks out of them. perhaps this will help \[log(a+b)=log(a)+log(b)\] \[log(a+b)=log(ba)=\frac{log(b)}{log(a)}\]
 one year ago

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

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

orbieBest ResponseYou've already chosen the best response.0
I understand the equations but I still can't find the right answers.
 one year ago

LolWolfBest ResponseYou've already chosen the best response.1
The first and last are the only two.
 one year ago

LolWolfBest ResponseYou've already chosen the best response.1
(By my above solution)
 one year ago

orbieBest 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.
 one year ago

LolWolfBest 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.
 one year ago

LolWolfBest ResponseYou've already chosen the best response.1
No, wait, never mind, it's only the first and the last.
 one year ago

LolWolfBest 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.
 one year ago

orbieBest ResponseYou've already chosen the best response.0
Thank you! It was the first and the last two that worked.
 one year ago

LolWolfBest 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.
 one year ago
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