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Thread: Log Equation

  1. #1
    Junior Member
    Joined
    Apr 2010
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    Log Equation

    $\displaystyle
    \ln x + \ln (x+2) = 4
    $

    The answer is:
    $\displaystyle
    x = -1 + \sqrt{1 + e^4}
    $

    so far I've done:

    $\displaystyle
    log_e x + log_e (x+2) = e^4
    $

    $\displaystyle
    log_e x^2+2x = e^4
    $

    Stuck.
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  2. #2
    Super Member
    Joined
    Jan 2008
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    Quote Originally Posted by desiderius1 View Post
    $\displaystyle
    \ln x + \ln (x+2) = 4
    $

    The answer is:
    $\displaystyle
    x = -1 + \sqrt{1 + e^4}
    $

    so far I've done:

    $\displaystyle
    log_e x + log_e (x+2) = e^4
    $ No it's not

    $\displaystyle
    log_e x^2+2x = e^4
    $ In the future please put brackets around x^2 + 2x

    Stuck.

    $\displaystyle \ln x + \ln (x+2) = 4 $

    $\displaystyle \ln | x(x+2)| = 4 $

    $\displaystyle x^2 + 2x = e^4 $

    $\displaystyle (x+1)^2 = e^4 + 1 $

    $\displaystyle x+1 = \sqrt {e^4 + 1} $

    $\displaystyle x = -1 + \sqrt{1+e^4} $
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  3. #3
    MHF Contributor

    Joined
    Apr 2005
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    Quote Originally Posted by desiderius1 View Post
    $\displaystyle
    \ln x + \ln (x+2) = 4
    $

    The answer is:
    $\displaystyle
    x = -1 + \sqrt{1 + e^4}
    $

    so far I've done:

    $\displaystyle
    log_e x + log_e (x+2) = e^4
    $
    That's NOT your orignal equation!

    $\displaystyle
    log_e x^2+2x = e^4
    $

    Stuck.
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