Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 766

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\[1)\log_{2}\sqrt[4]{2} = \log_{2}2^{\frac{1}{4}} = \frac{1}{4} = 0,25\]

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

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

\[= - 1\frac{1}{2} = - 1,5\]

\[3)\log_{0,5}\frac{1}{\sqrt{32}} = \log_{0,5}\left( \frac{1}{32} \right)^{\frac{1}{2}} =\]

\[= \log_{\frac{1}{2}}\left( \frac{1}{2} \right)^{\frac{5}{2}} = \frac{5}{2} = 2,5\]

\[4)\log_{7}\frac{\sqrt[3]{7}}{49} = \log_{7}\frac{7^{\frac{1}{3}}}{7^{2}} =\]

\[= \log_{7}7^{\frac{1}{3} - 2} = \frac{1}{3} - 2 = \frac{1}{3} - \frac{6}{3} =\]

\[= - \frac{5}{3} = - 1\frac{2}{3}\]

\[5)\log_{128}64 = \log_{2^{7}}2^{6} = \frac{6}{7}\]

\[6)\log_{27}243 = \log_{3^{3}}{3^{5} = \frac{5}{3}}\]

\[7)\log_{64}256 = \log_{4^{3}}4^{4} = \frac{4}{3}\]

\[8)\log_{81}27 = \log_{3^{4}}3^{3} = \frac{3}{4}\]

\[9)\log_{\sqrt{3}}\frac{1}{3\sqrt[4]{3}} = \log_{3^{\frac{1}{2}}}3^{- \frac{5}{4}} =\]

\[= 2 \cdot \left( - \frac{5}{4} \right) = - 2,5\]

\[10)\log_{\frac{1}{2}}\frac{1}{4 \cdot \sqrt[3]{2}} = \log_{\frac{1}{2}}\left( \frac{1}{2} \right)^{\frac{7}{3}} = \frac{7}{3}\]