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orbie

  • 3 years ago

Problem attached below.

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  1. orbie
    • 3 years ago
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  2. Outkast3r09
    • 3 years ago
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    if you look at all your answers...one sticks out of them. perhaps this will help \[log(a+b)=log(a)+log(b)\] \[log(-a+b)=log(b-a)=\frac{log(b)}{log(a)}\]

  3. LolWolf
    • 3 years ago
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    The only solutions to the system are: \[ \log(a+b)=\log(a)+\log(b)\implies\\ b=\frac{a}{a-1}, a\ne0, a-1\ne0 \]

  4. LolWolf
    • 3 years ago
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    @Outkast3r09 \( \log(a+b)=\log(a)+\log(b) \) Is *not* true for all cases.

  5. orbie
    • 3 years ago
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    I understand the equations but I still can't find the right answers.

  6. LolWolf
    • 3 years ago
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    The first and last are the only two.

  7. LolWolf
    • 3 years ago
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    (By my above solution)

  8. orbie
    • 3 years ago
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    mmm...that is not right either according to my math hw program. This problem is such a hassle.

  9. LolWolf
    • 3 years ago
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    Oh, I forgot to say that \(x>0\). The last two and the first are counter-examples to these.

  10. LolWolf
    • 3 years ago
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    No, wait, never mind, it's only the first and the last.

  11. LolWolf
    • 3 years ago
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    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.

  12. orbie
    • 3 years ago
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    Thank you! It was the first and the last two that worked.

  13. LolWolf
    • 3 years ago
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    Haha, yeah, I know why, now: the last one is imaginary, it *holds* but not in the real numbers.

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