# Expand the expression and write your answer without the exponents. \log_4

$$\displaystyle{{\log}_{{4}}{\frac{{{x}^{{2}}{y}^{{4}}}}{{{8}}}}}$$

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wornoutwomanC
Given:
$$\displaystyle{{\log}_{{4}}{\left({\frac{{{x}^{{2}}{y}^{{4}}}}{{{8}}}}\right)}}$$
To find the expression of $$\displaystyle{{\log}_{{4}}{\left({\frac{{{x}^{{2}}{y}^{{4}}}}{{{8}}}}\right)}}$$
Now applying the rule:
$$\displaystyle{{\log}_{{c}}{\left({\frac{{{a}}}{{{b}}}}\right)}}={{\log}_{{c}}{\left({a}\right)}}-{{\log}_{{c}}{\left({b}\right)}}$$
$$\displaystyle{{\log}_{{4}}{\left({\frac{{{x}^{{2}}{y}^{{4}}}}{{{8}}}}\right)}}={{\log}_{{4}}{\left({x}^{{2}}{y}^{{4}}\right)}}-{{\log}_{{4}}{\left({8}\right)}}$$
Now applying the rule in $$\displaystyle{{\log}_{{4}}{\left({x}^{{2}}{y}^{{4}}\right)}}$$
$$\displaystyle{{\log}_{{c}}{\left({a}{b}\right)}}={{\log}_{{c}}{\left({a}\right)}}+{{\log}_{{c}}{\left({b}\right)}}$$
$$\displaystyle={{\log}_{{4}}{\left({x}^{{2}}\right)}}+{{\log}_{{4}}{\left({y}^{{4}}\right)}}-{{\log}_{{4}}{\left({8}\right)}}$$
Now Applying the rule (3) in $$\displaystyle{{\log}_{{4}}{\left({x}^{{2}}\right)}}+{{\log}_{{4}}{\left({y}^{{4}}\right)}}$$
$$\displaystyle={2}{{\log}_{{4}}{\left({x}\right)}}+{4}{{\log}_{{4}}{\left({y}\right)}}-{{\log}_{{4}}{\left({8}\right)}}$$
$$\displaystyle={2}{{\log}_{{4}}{\left({x}\right)}}+{4}{{\log}_{{4}}{\left({y}\right)}}-{{\log}_{{{2}^{{2}}}}{\left({8}\right)}}$$
Now Applying the rule (3) in $$\displaystyle{{\log}_{{2}}{\left({8}\right)}}$$
$$\displaystyle{{\log}_{{a}}{\left({x}^{{b}}\right)}}={b}{{\log}_{{a}}{\left({x}\right)}}$$
$$\displaystyle={2}{{\log}_{{4}}{\left({x}\right)}}+{4}{{\log}_{{4}}{\left({y}\right)}}-{\frac{{{1}}}{{{2}}}}{3}{{\log}_{{2}}{\left({2}\right)}}$$
Now Applying the rule (5) in $$\displaystyle{{\log}_{{2}}{\left({2}\right)}}$$
$$\displaystyle{{\log}_{{a}}{\left({a}\right)}}={1}$$
$$\displaystyle={2}{{\log}_{{4}}{\left({x}\right)}}+{41}{{\log}_{{4}}{\left({y}\right)}}-{\frac{{{3}}}{{{2}}}}$$
Answer: $$\displaystyle{{\log}_{{4}}{\left({\frac{{{x}^{{2}}{y}^{{4}}}}{{{8}}}}\right)}}={2}{{\log}_{{4}}{\left({x}\right)}}+{4}{{\log}_{{4}}{\left({y}\right)}}={\frac{{{3}}}{{{2}}}}$$