[latex]1)\ log_{frac{1}{4}}frac{2sqrt[5]{64}}{sqrt{8}}=log_{frac{1}{4}}frac{2cdot 2sqrt[5]{2}}{2sqrt{2}}=log_{frac{1}{4}}frac{2sqrt[5]{2}}{sqrt{2}}=log_{frac{1}{4}}(sqrt[5]{2}}cdot sqrt{2})=\ =log_{frac{1}{4}}(sqrt[5]{2}})+log_{frac{1}{4}}(sqrt{2})=log_{2^{-2}}(sqrt[5]{2}})+log_{2^{-2}}(sqrt{2}})=\ =frac{-1}{2}log_2(sqrt[5]{2}})-frac{1}{2}log_2(sqrt{2}})=\ =frac{-1}{2}log_22^{frac{1}{5}}-frac{1}{2}log_22^{frac{1}{2}}=\ [/latex] [latex]=frac{-1}{2}cdot frac{1}{5}-frac{1}{2}cdot frac{1}{2}=frac{-1}{10}-frac{1}{4}=frac{-2}{20}-frac{5}{20}=frac{-7}{20}[/latex] [latex]2)\ 2log_{frac{1}{4}}8-log_{sqrt{3}}9=log_{frac{1}{4}}8^2-log_{sqrt{3}}(sqrt{3})^4=\ =log_{frac{1}{4}}64-4log_{sqrt{3}}(sqrt{3})=log_{frac{1}{4}}4^{3}-4log_{sqrt{3}}(sqrt{3})=\ =log_{frac{1}{4}}(frac{1}{4})^{-3}-4log_{sqrt{3}}(sqrt{3}) =-3-4=-7[/latex]
Rozwiązanie w załączniku.
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