Can someone show me how this is possible?
$\displaystyle
\ln \left( {\frac{{\sqrt 2 + 1}}
{{\sqrt 2 - 1}}} \right) = 2\ln \left( {\sqrt 2 + 1} \right)
$
Thank you.
Multiply the inside of the natural log by the conjugate of the denominator:
$\displaystyle
\ln \left( {\frac{{\sqrt 2 + 1}}
{{\sqrt 2 - 1}}}\cdot{\color{red}\frac{\sqrt 2+1}{\sqrt 2+1}} \right)=\ln\left(\frac{(\sqrt 2+1)^2}{2-1}\right)=\ln\left( (\sqrt 2+1)^2\right)=\color{red}2\ln\left(\sqrt 2+1\right)$
Does this make sense?
--Chris