Answer
See proof
Work Step by Step
$\ln (\frac{e^{x}}{e^{y}})=\ln (e^{x})-\ln (e^{y})= x-y=\ln (e^{x-y})$
Since $\ln$ is a one-to-one function, it follows that $\frac{e^{x}}{e^{y}}=e^{x-y}$.
In the above proof, we have used two relations (or laws) which are
$\ln (e^{x})=x$
and
$\ln(\frac{x}{y})=\ln x-\ln y$.
