Results 1 to 2 of 2

Thread: I need help once more

  1. #1
    Newbie
    Joined
    May 2007
    Posts
    3

    I need help once more

    Make x the subject of the formula

    L=2K[1 + ln(x/a)]
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Jonboy's Avatar
    Joined
    May 2007
    From
    South Point
    Posts
    193
    $\displaystyle L\,=\,2K[1\,+\,ln(\frac{x}{a})]$

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

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

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

    Cross multiply:..$\displaystyle \boxed{x\,=\,e^{\frac{L}{2K}\,-\,1}\,\cdot\,a}$
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum