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Math Help - I need help once more

  1. #1
    Newbie
    Joined
    May 2007
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    3

    I need help once more

    Make x the subject of the formula

    L=2K[1 + ln(x/a)]
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  2. #2
    Member Jonboy's Avatar
    Joined
    May 2007
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    South Point
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    L\,=\,2K[1\,+\,ln(\frac{x}{a})]

    Divide by 2K:.. \frac{L}{2K}\,=\,1\,+\,ln(\frac{x}{a})

    Subtract 1:.. \frac{L}{2K}\,-\,1\,=\,ln(\frac{x}{a})

    Raise both to a base of e:..  e^{\frac{L}{2K}\,-\,1}\,=\,\frac{x}{a}

    Cross multiply:.. \boxed{x\,=\,e^{\frac{L}{2K}\,-\,1}\,\cdot\,a}
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