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Math Help - [SOLVED] solving a log

  1. #1
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    [SOLVED] solving a log

    I'm having difficulties solving this log. I was hoping someone could point me in the right direction
    lnx +ln(x+4) = 2
    lx(x(x+4)) = 2
    x(x+4) = e^2

    I could write that out to a sorta-polynomial of
    x^2 + x - e^2
    but I'm not really sure where I should be proceeding from here. Thanks in advance for any assistance.
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  2. #2
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    Use the Quadratic Formula with a = 1, b = 1, and c = -e^2.
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  3. #3
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    I'd tried that route by my answer was way off from what the book had. Let me do the work and perhaps you could point out what I'm doing wrong?

    x^2+x-e^2
    into the quadratic equation
    \frac{-1 + \sqrt{1+4(e^2)}}{2}
    \frac{-1 + \sqrt{5e^2}}{2}

    If I solve that I get a positive answer of 2.539 (roughly) and a negative answer which is invalid. Unfortunately the book has

    1 + \sqrt{1+e} \approx 2.928
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  4. #4
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    I believe the error lies in your conversion of x(x + 4) = e^2 to x^2 + x - e^2 = 0, specifically your distribution of x to 4.
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  5. #5
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    Quote Originally Posted by satis View Post
    I'm having difficulties solving this log. I was hoping someone could point me in the right direction
    lnx +ln(x+4) = 2
    lx(x(x+4)) = 2
    x(x+4) = e^2

    I could write that out to a sorta-polynomial of
    x^2 + x - e^2
    but I'm not really sure where I should be proceeding from here. Thanks in advance for any assistance.
    You distributed x incorrectly in that last step, try using the quadratic formula on:
    x^2+4x-e^2
    Or, you could factor x(x+4)-e^2=0 fairly easily.
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  6. #6
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    sure enough... for some reason I kept reading that 4 as a 1. Repeatedly. Thanks for pointing out the issue guys.
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