Given \(\displaystyle{{\log}_{{{2}}}{\left({\frac{{{x}{y}^{{{4}}}}}{{{z}^{{{5}}}}}}\right)}}\)

Here we write as the sum and\or difference of logarithms.

Now \(\displaystyle{{\log}_{{{2}}}{\left({\frac{{{x}{y}^{{{4}}}}}{{{z}^{{{5}}}}}}\right)}}\)

\(\displaystyle={{\log}_{{{2}}}{\left({x}{y}^{{{4}}}\right)}}-{{\log}_{{{2}}}{\left({z}^{{{5}}}\right)}}\)

\(\displaystyle={\left({{{\log}_{{{2}}}^{{{x}}}}+}{{\log}_{{{2}}}^{{{y}{4}}}}\right)}-{{\log}_{{{2}}}{\left({z}^{{{5}}}\right)}}{\left[\because{{{\log}_{{{a}}}^{{{\left({m}{n}\right)}}}}=}{{{\log}_{{{a}}}^{{{m}}}}+}{{\log}_{{{a}}}^{{{n}}}}\right]}\)

\(\displaystyle={{{\log}_{{{2}}}^{{{x}}}}+}{4}{{{\log}_{{{2}}}^{{{y}}}}-}{5}{{{\log}_{{{2}}}^{{{z}}}}{\left[\because{{{\log}_{{{n}}}^{{{x}{m}}}}=}{m}{{\log}_{{{a}}}^{{{x}}}}\right]}}\)

Hence \(\displaystyle{{\log}_{{{2}}}{\left({\frac{{{x}{y}^{{{4}}}}}{{{z}^{{{5}}}}}}\right)}}={{{\log}_{{{2}}}^{{{x}}}}+}{4}{{{\log}_{{{2}}}^{{{y}}}}-}{5}{{{\log}_{{{2}}}^{{{2}}}}}\)

Here we write as the sum and\or difference of logarithms.

Now \(\displaystyle{{\log}_{{{2}}}{\left({\frac{{{x}{y}^{{{4}}}}}{{{z}^{{{5}}}}}}\right)}}\)

\(\displaystyle={{\log}_{{{2}}}{\left({x}{y}^{{{4}}}\right)}}-{{\log}_{{{2}}}{\left({z}^{{{5}}}\right)}}\)

\(\displaystyle={\left({{{\log}_{{{2}}}^{{{x}}}}+}{{\log}_{{{2}}}^{{{y}{4}}}}\right)}-{{\log}_{{{2}}}{\left({z}^{{{5}}}\right)}}{\left[\because{{{\log}_{{{a}}}^{{{\left({m}{n}\right)}}}}=}{{{\log}_{{{a}}}^{{{m}}}}+}{{\log}_{{{a}}}^{{{n}}}}\right]}\)

\(\displaystyle={{{\log}_{{{2}}}^{{{x}}}}+}{4}{{{\log}_{{{2}}}^{{{y}}}}-}{5}{{{\log}_{{{2}}}^{{{z}}}}{\left[\because{{{\log}_{{{n}}}^{{{x}{m}}}}=}{m}{{\log}_{{{a}}}^{{{x}}}}\right]}}\)

Hence \(\displaystyle{{\log}_{{{2}}}{\left({\frac{{{x}{y}^{{{4}}}}}{{{z}^{{{5}}}}}}\right)}}={{{\log}_{{{2}}}^{{{x}}}}+}{4}{{{\log}_{{{2}}}^{{{y}}}}-}{5}{{{\log}_{{{2}}}^{{{2}}}}}\)