Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1273

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\[1)\ \log_{4}{\sin\frac{\pi}{4}}\mathbf{=}\log_{4}\frac{1}{\sqrt{2}} =\]

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

\[2)\ \log_{10}{\text{tg}\frac{\pi}{4}} = \log_{10}1 =\]

\[= \log_{10}10^{0} = 0;\]

\[3)\ \log_{8}{\sin\frac{3\pi}{4}} = \log_{8}{\sin\left( \pi - \frac{\pi}{4} \right)} =\]

\[= \log_{8}{\sin\frac{\pi}{4}} = \log_{8}\frac{1}{\sqrt{2}} =\]

\[= \log_{2^{3}}2^{- \frac{1}{2}} = \log_{2^{3}}\left( 2^{3} \right)^{- \frac{1}{6}} = - \frac{1}{6};\]

\[4)\ \log_{2}{\cos\frac{\pi}{3}} = \log_{2}\frac{1}{2} =\]

\[= \log_{2}2^{- 1} = - 1;\]

\[5)\ \log_{3}1 - \log_{4}{\text{tg}\frac{\pi}{4}} \bullet \log_{5}{\cos 0} =\]

\[= \log_{3}3^{0} - \log_{4}1 \bullet \log_{5}1 =\]

\[= 0 - \log_{4}4^{0} \bullet \log_{5}5^{0} = 0.\]