anonymous
  • anonymous
Problem attached below.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
if you look at all your answers...one sticks out of them. perhaps this will help \[log(a+b)=log(a)+log(b)\] \[log(-a+b)=log(b-a)=\frac{log(b)}{log(a)}\]
anonymous
  • anonymous
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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anonymous
  • anonymous
@Outkast3r09 \( \log(a+b)=\log(a)+\log(b) \) Is *not* true for all cases.
anonymous
  • anonymous
I understand the equations but I still can't find the right answers.
anonymous
  • anonymous
The first and last are the only two.
anonymous
  • anonymous
(By my above solution)
anonymous
  • anonymous
mmm...that is not right either according to my math hw program. This problem is such a hassle.
anonymous
  • anonymous
Oh, I forgot to say that \(x>0\). The last two and the first are counter-examples to these.
anonymous
  • anonymous
No, wait, never mind, it's only the first and the last.
anonymous
  • anonymous
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.
anonymous
  • anonymous
Thank you! It was the first and the last two that worked.
anonymous
  • anonymous
Haha, yeah, I know why, now: the last one is imaginary, it *holds* but not in the real numbers.

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