# log problem I am struggling with

• Dec 11th 2007, 09:46 AM
richard_c
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
• Dec 11th 2007, 12:26 PM
earboth
Quote:

Originally Posted by richard_c
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.