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:
$\displaystyle
R=\dfrac{\ln\left(\frac{2L}{D} \right) \cdot \left(L+ \dfrac{\ln\left(\frac{L}{2Z}\right) }{\ln\left(\frac{2L}{D} \right)} \right)}{2 \pi k L}$
First multiply by $\displaystyle 2 \pi k L$. Afterwards expand the bracket at the RHS of the equation. You'll get:
$\displaystyle 2 \pi k L \cdot R=L \cdot \ln\left(\frac{2L}{D} \right) + \ln\left(\frac{L}{2Z} \right)$
Subtract the first summand of the RHS of the equation:
$\displaystyle 2 \pi k L \cdot R- L \cdot \ln\left(\frac{2L}{D} \right) = \ln(L) - \ln(2Z)$
$\displaystyle 2 \pi k L \cdot R- L \cdot \ln\left(\frac{2L}{D} \right) - \ln(L) = \ln(2) + \ln(Z)$
$\displaystyle 2 \pi k L \cdot R- L \cdot \ln\left(\frac{2L}{D} \right) - \ln(L) - \ln(2) = \ln(Z)$
And now de-logarithmize(?):
$\displaystyle e^{2 \pi k L \cdot R- L \cdot \ln\left(\frac{2L}{D} \right) - \ln(L) - \ln(2)} = e^{\ln(Z)} = Z$
And now try to simplify this nightmare of a solution.