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Math Help - log problem I am struggling with

  1. #1
    Newbie richard_c's Avatar
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    log problem I am struggling with

    Having real pains too with this one.

    Please see attachement - cant use the equation editior on here

    Sorry.

    RC
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  2. #2
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    earboth's Avatar
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    Quote Originally Posted by richard_c View Post
    Having real pains too with this one.

    Please see attachement - cant use the equation editior on here

    Sorry.

    RC
    Hello,

    I'm going to isolate Z. I assume that l = L and that ln() = Ln(). You have:
    <br />
R=\dfrac{\ln\left(\frac{2L}{D} \right) \cdot \left(L+ \dfrac{\ln\left(\frac{L}{2Z}\right) }{\ln\left(\frac{2L}{D} \right)}  \right)}{2 \pi k L}

    First multiply by 2 \pi k L. Afterwards expand the bracket at the RHS of the equation. You'll get:

    2 \pi k L \cdot R=L \cdot \ln\left(\frac{2L}{D} \right) + \ln\left(\frac{L}{2Z} \right)

    Subtract the first summand of the RHS of the equation:

    2 \pi k L \cdot R- L \cdot \ln\left(\frac{2L}{D} \right) = \ln(L) - \ln(2Z)

    2 \pi k L \cdot R- L \cdot \ln\left(\frac{2L}{D} \right) - \ln(L) = \ln(2) + \ln(Z)

    2 \pi k L \cdot R- L \cdot \ln\left(\frac{2L}{D} \right) - \ln(L) - \ln(2) = \ln(Z)

    And now de-logarithmize(?):

    e^{2 \pi k L \cdot R- L \cdot \ln\left(\frac{2L}{D} \right) - \ln(L) - \ln(2)} = e^{\ln(Z)} = Z

    And now try to simplify this nightmare of a solution.
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