Answer
$ln(\frac{x^{2}}{y^{\frac{1}{2}}})$
Work Step by Step
According to the power rule of logarithms, we know that $log_{b}M^{p}=plog_{b}M$ (when $b$ and $M$ are positive real numbers, $b\ne1$, and $p$ is any real number).
Therefore, $2ln(x)-\frac{1}{2}ln(y)=ln(x^{2})-ln(y^{\frac{1}{2}})$.
Based on the quotient rule of logarithms, we know that $log_{b}(\frac{M}{N})=log_{b}M-log_{b}N$ (where $b$, $M$, and $N$ are positive real numbers and $b\ne1$).
Therefore, $ ln(x^{2})-ln(y^{\frac{1}{2}})=ln(\frac{x^{2}}{y^{\frac{1}{2}}})$.
In this case, the given logarithm is a natural logarithm with an understood base of $e$.