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Mathematics
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if you look at all your answers...one sticks out of them. perhaps this will help \[log(a+b)=log(a)+log(b)\] \[log(-a+b)=log(b-a)=\frac{log(b)}{log(a)}\]
The only solutions to the system are: \[ \log(a+b)=\log(a)+\log(b)\implies\\ b=\frac{a}{a-1}, a\ne0, a-1\ne0 \]

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@Outkast3r09 \( \log(a+b)=\log(a)+\log(b) \) Is *not* true for all cases.
I understand the equations but I still can't find the right answers.
The first and last are the only two.
(By my above solution)
mmm...that is not right either according to my math hw program. This problem is such a hassle.
Oh, I forgot to say that \(x>0\). The last two and the first are counter-examples to these.
No, wait, never mind, it's only the first and the last.
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.
Thank you! It was the first and the last two that worked.
Haha, yeah, I know why, now: the last one is imaginary, it *holds* but not in the real numbers.

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