Answer
Write the right side of the equation as a single logarithm expression.
Refer to the step-by-step by to see the steps.
Work Step by Step
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,
$\ln{(1+e^{2x})}=2x+\ln{(1+e^{-2x})}$
