## Precalculus (10th Edition)

We know that if $e^x=y$, then $\ln {y}=x$, Hence, $2x=\ln{(e^{2x})}$ Thus, the right side of the equation is equivalent to $=\ln{e^{2x}}+\ln{(1+e^{-2x})}$ We know that $\log_a {x}+\log_a {y}=\log_a {(xy)}$. Hence, the right side of the given equation becomes $2x+\ln{(1+e^{-2x})}\\ =\ln{(e^{2x})}+\ln{(1+e^{-2x})}\\ =\ln{[e^{2x}\cdot(1+e^{-2x})]}\\ =\ln{(e^{2x}+e^{2x-2x})}\\ =\ln{(e^{2x}+e^0)}\\ =\ln{(e^{2x}+1)}\\ =\ln{(1+e^{2x})}$ Therefore, $\ln{(1+e^{2x})}=2x+\ln{(1+e^{-2x})}$