# Write as the sum and\or difference of logarithms. Express powers as factors. \log_{2}(\frac{xy^{4}}{z^{5}})

Write as the sum and\or difference of logarithms. Express powers as factors.
$$\displaystyle{{\log}_{{{2}}}{\left({\frac{{{x}{y}^{{{4}}}}}{{{z}^{{{5}}}}}}\right)}}$$

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