Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 384

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

\[= 2\log_{3}\frac{1}{3^{\frac{4}{3}}} = 2\log_{3}3^{- \frac{4}{3}} =\]

\[= 2 \bullet \left( - \frac{4}{3} \right) = - \frac{8}{3}\]

\[2)\log_{\sqrt{5}}\frac{1}{25\sqrt[4]{5}} = \log_{5^{\frac{1}{2}}}\frac{1}{5^{2} \bullet 5^{\frac{1}{4}}} =\]

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

\[= 2 \bullet \left( - \frac{9}{4} \right) = - \frac{9}{2}\]

\[3)\ 2^{2 - \log_{2}5} = \frac{2^{2}}{2^{\log_{2}5}} = \frac{4}{5} = 0,8\]

\[4)\ {3,6}^{\log_{3,6}10 + 1} =\]

\[= {3,6}^{\log_{3,6}10} \bullet 3,6 = 10 \bullet 3,6 = 36\]

\[5)\ 2\log_{5}\sqrt{5} + 3\log_{2}8 =\]

\[= \log_{5}\left( \sqrt{5} \right)^{2} + \log_{2}8^{3} =\]

\[= \log_{5}5 + \log_{2}512 =\]

\[= 1 + \log_{2}2^{9} = 1 + 9 = 10\]

\[6)\log_{2}{\log_{2}{\log_{2}2^{16}}} =\]

\[= \log_{2}{\log_{2}16} = \log_{2}{\log_{2}2^{4}} =\]

\[= \log_{2}4 = \log_{2}2^{2} = 2\]