Question

How is this true? \[\ln(e^{x}+e^{-x})=\ln(e^{2x}+1)-x\]

Answer 1

\begin{eqnarray*}\log(e^x + e^{-x}) &=& \log\left(e^x + \frac{1}{e^x}\right) \\&=& \log\left(\frac{e^x \cdot e^x + 1}{e^x}\right)\\&=&\log\left(\frac{e^{2x} + 1}{e^x}\right) \\&=& \log(e^{2x} + 1) - \log(e^x) \\&=&\log(e^{2x} + 1) - x.\end{eqnarray*}

Answer 2

e^x + e^-x = (1 + e^2x) / e^x
ln (1 + e^2x) / e^x
=ln(1 + e^2x) - ln e^x
=ln (1 + e^2x) - x