Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 296

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\[1)\ \frac{\log_{2}24 - \frac{1}{2}\log_{2}72}{\log_{3}18 - \frac{1}{3}\log_{3}72} =\]

\[= \frac{\log_{2}24 - \log_{2}\sqrt{72}}{\log_{3}18 - \log_{3}\sqrt[3]{72}} =\]

\[= \frac{\log_{2}\frac{24}{\sqrt{72}}}{\log_{3}\frac{18}{\sqrt[3]{72}}} = \frac{\log_{2}\sqrt{\frac{24^{2}}{24 \bullet 3}}}{\log_{3}\sqrt[3]{\frac{18^{3}}{18 \bullet 4}}} =\]

\[= \frac{\log_{2}\sqrt{\frac{24}{3}}}{\log_{3}\sqrt[3]{\frac{324}{4}}} = \frac{\log_{2}\sqrt{8}}{\log_{3}\sqrt[3]{81}} =\]

\[= \frac{\log_{2}2^{\frac{3}{2}}}{\log_{3}3^{\frac{4}{3}}} = \frac{3}{2}\ :\frac{4}{3} = \frac{3}{2} \bullet \frac{3}{4} =\]

\[= \frac{9}{8} = 1\frac{1}{8}\]

\[2)\ \frac{\log_{7}14 - \frac{1}{3}\log_{7}56}{\log_{6}30 - \frac{1}{2}\log_{6}150} =\]

\[= \frac{\log_{7}14 - \log_{7}\sqrt[3]{56}}{\log_{6}30 - \log_{6}\sqrt{150}} =\]

\[= \frac{\log_{7}\frac{14}{\sqrt[3]{56}}}{\log_{6}\frac{30}{\sqrt{150}}} = \frac{\log_{7}\sqrt[3]{\frac{14^{3}}{14 \bullet 4}}}{\log_{6}\sqrt{\frac{30^{2}}{30 \bullet 5}}} =\]

\[= \frac{\log_{7}\sqrt[3]{\frac{196}{4}}}{\log_{6}\sqrt{\frac{30}{5}}} = \frac{\log_{7}\sqrt[3]{49}}{\log_{6}\sqrt{6}} =\]

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

\[= \frac{4}{3} = 1\frac{1}{3}\]

\[3)\ \frac{\log_{2}4 + \log_{2}\sqrt{10}}{\log_{2}20 + 3\log_{2}2} =\]

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

\[= \frac{\log_{2}2^{2} + \frac{1}{2}\left( \log_{2}2 + \log_{2}5 \right)}{\log_{2}4 + \log_{2}5 + 3} =\]

\[= \frac{2 + \frac{1}{2}\left( \log_{2}2 + \log_{2}5 \right)}{\log_{2}2^{2} + \log_{2}5 + 3} =\]

\[= \frac{\frac{1}{2}\left( 4 + 1 + \log_{2}5 \right)}{2 + \log_{2}5 + 3} =\]

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

\[4)\ \frac{3\log_{7}2 - \frac{1}{2}\log_{7}64}{4\log_{5}2 + \frac{1}{3}\log_{5}27} =\]

\[= \frac{\log_{7}2^{3} - \log_{7}\sqrt{64}}{\log_{5}2^{4} + \log_{5}\sqrt[3]{27}} =\]

\[= \frac{\log_{7}8 - \log_{7}8}{\log_{5}16 + \log_{5}3} = 0\]