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Math Help - Log equation question...

  1. #1
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    Log equation question...

    Hi all. I'm having difficulty seeing how my solution to a problem is wrong. Here it is, along with my attempt:

    log(2x+1) = 1-log(x-1)
    log(2x+1) + log(x-1) = 1
    log((2x+1)(x-1)) = 1
    log(2x^2-x-1) = 1
    2x^2-x-1 = 10
    (2x+1)(x-1)=10
    2x=9, x=9/2
    x=11

    I'm sure it's a case of "oh, duh." :P Thanks in advance...
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  2. #2
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    Quote Originally Posted by NJDesigner View Post
    Hi all. I'm having difficulty seeing how my solution to a problem is wrong. Here it is, along with my attempt:

    log(2x+1) = 1-log(x-1) <-- don't forget: x > 1
    log(2x+1) + log(x-1) = 1
    log((2x+1)(x-1)) = 1
    log(2x^2-x-1) = 1
    2x^2-x-1 = 10
    ...
    I'm sure it's a case of "oh, duh." :P Thanks in advance...
    This is a quadratic equation. To solve it use the quadratic formula:

    2x^2-x-11=0~\implies~x=\dfrac{1 \pm \sqrt{1^2-4 \cdot 2 \cdot (-11)}}{2 \cdot 2}

    You'll get 2 solutions but only one satisfies the original equation.
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  3. #3
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    A- ha!

    Thanks! I had a feeling actually but I wasn't confident about it. I'm looking how to give points now (newbie here)
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