Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 292

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\[1)\log_{13}\sqrt[5]{169} = \log_{13}169^{\frac{1}{5}} =\]

\[= \frac{1}{5}\log_{13}169 = \frac{1}{5}\log_{13}13^{2} =\]

\[= \frac{1}{5} \bullet 2 = 0,4\]

\[2)\log_{11}\sqrt[3]{121} = \log_{11}121^{\frac{1}{3}} =\]

\[= \frac{1}{3}\log_{11}121 = \frac{1}{3}\log_{11}11^{2} =\]

\[= \frac{1}{3} \bullet 2 = \frac{2}{3}\]

\[3)\log_{\frac{1}{3}}\sqrt[4]{243} = \log_{\frac{1}{3}}243^{\frac{1}{4}} =\]

\[= \frac{1}{4}\log_{\frac{1}{3}}243 = \frac{1}{4}\log_{\frac{1}{3}}3^{5} =\]

\[= \frac{1}{4}\log_{\frac{1}{3}}\left( \frac{1}{3} \right)^{- 5} = \frac{1}{4} \bullet - 5 = - 1,25\]

\[4)\log_{2}\frac{1}{\sqrt[6]{128}} = \log_{2}\frac{1}{128^{\frac{1}{6}}} =\]

\[= \log_{2}128^{- \frac{1}{6}} = - \frac{1}{6}\log_{2}128 =\]

\[= - \frac{1}{6}\log_{2}2^{7} = - \frac{1}{6} \bullet 7 = - 1\frac{1}{6}\]