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Math Help - Tricky (ln) problem

  1. #1
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    Tricky (ln) problem

    I was assigned the following:

    The only solution of the equation ln(x) + ln(x-2) =1 is x = ?

    I am unsure how to solve for x.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Tricky (ln) problem

    Use the fact that \ln(a)+\ln(b)=\ln(a\cdot b) and \ln(e)=1 therefore the equation can be arranged as:
    \ln[x(x-2)]=\ln(e)
    \Leftrightarrow x(x-2)=e

    Solve this quadratic equation.
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  3. #3
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    Re: Tricky (ln) problem

    I solved for x, and came up with 1 + sqrt(4 +4e). However, this is still wrong, and I am unsure why!
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  4. #4
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    e^(i*pi)'s Avatar
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    Re: Tricky (ln) problem

    Quote Originally Posted by Dmiller93 View Post
    I solved for x, and came up with 1 + sqrt(4 +4e). However, this is still wrong, and I am unsure why!
    Following on from Siron's working:

    x^2-2x = e \Leftrightarrow x^2-2x-e = 0

    Since e is a number solve using the quadratic formula:

    x = \dfrac{2\pm \sqrt{4+4e}}{2} = \dfrac{2 \pm \sqrt{4(1+e)}}{2} = \dfrac{2 \pm \sqrt{4}\sqrt{1+e}}{2}

    = \dfrac{2 \pm 2\sqrt{1+e}}{2} = \dfrac{2(1 \pm \sqrt{1+e})}{2} =1 \pm \sqrt{1+e}


    Since \sqrt{1+e} > 1 discard the negative solution due to domain issues (x>2)

    x = 1 + \sqrt{1+e}
    Last edited by e^(i*pi); September 23rd 2011 at 02:07 PM. Reason: sign error
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