$\begin{align} & \text{Given }a={{x}^{2n}},\text{ }b={{y}^{-n}},\text{ }c={{x}^{\tfrac{n}{2}}}{{y}^{\tfrac{2}{3}}} \\ & i)abc={{x}^{2n}}\times {{y}^{-n}}\times {{x}^{\tfrac{n}{2}}}{{y}^{\tfrac{2}{3}}}={{x}^{2n+\tfrac{n}{2}}}{{y}^{-n+\tfrac{2}{3}}}={{x}^{\tfrac{5n}{2}}}{{y}^{\tfrac{2-3n}{3}}} \\ & ii)\frac{ab}{c}=\frac{{{x}^{2n}}\times {{y}^{-n}}}{{{x}^{\tfrac{n}{2}}}{{y}^{\tfrac{2}{3}}}}={{x}^{2n-\tfrac{n}{2}}}{{y}^{-n-\tfrac{2}{3}}}={{x}^{\tfrac{3n}{2}}}{{y}^{\tfrac{-3n-2}{3}}}=\frac{{{x}^{\tfrac{3n}{2}}}}{{{y}^{\tfrac{3n-2}{3}}}} \\ & iii){{\log }_{e}}a+{{\log }_{e}}bc={{\log }_{e}}{{x}^{2n}}+{{\log }_{e}}\left( {{y}^{-n}}{{x}^{\tfrac{n}{2}}}{{y}^{\tfrac{2}{3}}} \right)={{\log }_{e}}\left( {{x}^{2n}}\times {{y}^{-n}}{{x}^{\tfrac{n}{2}}}{{y}^{\tfrac{2}{3}}} \right) \\ & {{\log }_{e}}a+{{\log }_{e}}bc={{\log }_{e}}\left( {{x}^{2n+\tfrac{n}{2}}}{{y}^{-n+\tfrac{2}{3}}} \right)={{\log }_{e}}\left( {{x}^{\tfrac{5n}{2}}}{{y}^{\tfrac{2-3n}{3}}} \right)={{\log }_{e}}{{x}^{\tfrac{5n}{2}}}+{{\log }_{e}}{{y}^{\tfrac{2-3n}{3}}} \\ & {{\log }_{e}}a+{{\log }_{e}}bc=\tfrac{5n}{2}{{\log }_{e}}x+\tfrac{2-3n}{3}{{\log }_{e}}y \\ & iv){{\log }_{e}}\frac{b}{c}-{{\log }_{e}}c={{\log }_{e}}\frac{{{y}^{-n}}}{{{x}^{\tfrac{n}{2}}}{{y}^{\tfrac{2}{3}}}}-{{\log }_{e}}{{x}^{\tfrac{n}{2}}}{{y}^{\tfrac{2}{3}}} \\ & {{\log }_{e}}\frac{b}{c}-{{\log }_{e}}c={{\log }_{e}}\frac{\tfrac{{{y}^{-n}}}{{{x}^{\tfrac{n}{2}}}{{y}^{\tfrac{2}{3}}}}}{{{x}^{\tfrac{n}{2}}}{{y}^{\tfrac{2}{3}}}}=\log e\frac{{{y}^{-n}}}{({{x}^{\tfrac{n}{2}}}{{y}^{\tfrac{2}{3}}})({{x}^{\tfrac{n}{2}}}{{y}^{\tfrac{2}{3}}})}=\log e\frac{{{y}^{-n}}}{{{({{x}^{\tfrac{n}{2}}}{{y}^{\tfrac{2}{3}}})}^{2}}} \\ & {{\log }_{e}}\frac{b}{c}-{{\log }_{e}}c={{\log }_{e}}{{y}^{-n}}-{{\log }_{e}}{{({{x}^{\tfrac{n}{2}}}{{y}^{\tfrac{2}{3}}})}^{2}}={{\log }_{e}}{{y}^{-n}}-{{\log }_{e}}({{x}^{n}}{{y}^{\tfrac{4}{3}}}) \\ & {{\log }_{e}}\frac{b}{c}-{{\log }_{e}}c={{\log }_{e}}\left( \frac{{{y}^{-n}}}{{{x}^{n}}{{y}^{\tfrac{4}{3}}}} \right)={{\log }_{e}}\left( \frac{{{y}^{-n-\tfrac{4}{3}}}}{{{x}^{n}}} \right) \\ & {{\log }_{e}}\frac{b}{c}-{{\log }_{e}}c~={{\log }_{e}}\left( \frac{{{y}^{\tfrac{-3n-4}{3}}}}{{{x}^{n}}} \right)={{\log }_{e}}\left( \frac{{{y}^{\tfrac{^{-(3n+4)}}{3}}}}{{{x}^{n}}} \right) \\ & {{\log }_{e}}\frac{b}{c}-{{\log }_{e}}c={{\log }_{e}}\left[ {{x}^{n}}{{y}^{\tfrac{^{-(3n+4)}}{3}}} \right]=-{{\log }_{e}}\left[ {{x}^{n}}{{y}^{\tfrac{^{(3n+4)}}{3}}} \right] \\ & vi)\frac{{{\log }_{e}}a-{{\log }_{c}}b}{{{\log }_{e}}c}=\frac{{{\log }_{e}}{{x}^{2n}}-{{\log }_{e}}{{y}^{-n}}}{{{\log }_{e}}{{x}^{\tfrac{n}{2}}}{{y}^{\tfrac{2}{3}}}}=\frac{{{\log }_{e}}\frac{{{x}^{2n}}}{{{y}^{-n}}}}{{{\log }_{e}}{{x}^{\tfrac{n}{2}}}{{y}^{\tfrac{2}{3}}}} \\ & \frac{{{\log }_{e}}a-{{\log }_{c}}b}{{{\log }_{e}}c}=\frac{\log {{x}^{2n}}{{y}^{n}}}{{{\log }_{e}}{{x}^{\tfrac{n}{2}}}{{y}^{\tfrac{2}{3}}}} \\ \end{align}$