Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 768

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\[1)\log_{2}{\log_{3}81} = \log_{2}{\log_{3}3^{4}} =\]

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

\[2)\log_{3}{\log_{2}8} = \log_{3}{\log_{2}2^{3}} =\]

\[= \log_{3}3 = 1\]

\[3)\ 2\log_{27}{\log_{10}1000} =\]

\[= 2\log_{27}{\log_{10}10^{3}} =\]

\[= 2\log_{27}3 = 2\log_{27}\left( 3^{3} \right)^{\frac{1}{3}} =\]

\[= 2\log_{27}27^{\frac{1}{3}} = 2 \bullet \frac{1}{3} = \frac{2}{3}\]

\[4)\ \frac{1}{3}\log_{9}{\log_{2}8} =\]

\[= \frac{1}{3}\log_{9}{\log_{2}2^{3}} = \frac{1}{3}\log_{9}3 =\]

\[= \frac{1}{3}\log_{9}\left( 3^{2} \right)^{\frac{1}{2}} = \frac{1}{3}\log_{9}9^{\frac{1}{2}} =\]

\[= \frac{1}{3} \bullet \frac{1}{2} = \frac{1}{6}\]

\[5)\ 3\log_{2}{\log_{4}16 + \log_{\frac{1}{2}}2} =\]

\[= 3\log_{2}{\log_{4}4^{2}} + \log_{\frac{1}{2}}\left( \frac{1}{2} \right)^{- 1} =\]

\[= 3\log_{2}2 + ( - 1) =\]

\[= 3 \bullet 1 - 1 = 3 - 1 = 2\]

\[6)\ 2\log_{4}{\log_{16}256} + \log_{\sqrt{2}}8 =\]

\[= 2\log_{4}{\log_{16}16^{2}} + \log_{2^{\frac{1}{2}}}2^{3} =\]

\[= 2\log_{4}2 + 2 \cdot 3\log_{2}2 =\]

\[= 2\log_{2^{2}}2 + 6 =\]

\[= 2 \cdot \frac{1}{2}\log_{2}2 + 6 = 1 + 6 = 7\]