Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 811

\[\boxed{\mathbf{811}.}\]

\[1)\ \frac{\log_{3}216}{\log_{8}3} - \frac{\log_{3}24}{\log_{72}3} =\]

\[= \log_{3}\left( 3^{3} \cdot 2^{3} \right) \cdot \log_{3}2^{3} -\]

\[- \log_{3}\left( 2^{3} \cdot 3 \right) \cdot \log_{3}\left( 2^{3} \cdot 3^{2} \right) =\]

\[= \left( 3 + \log_{3}8 \right) \cdot \log_{3}8 -\]

\[- \left( \log_{3}8 + 1 \right)\left( \log_{3}8 + 2 \right) =\]

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

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

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

\[- \left( \log_{3}8 \right)^{2} - 2\log_{3}8 -\]

\[- \log_{3}8 - 2 = - 2.\]

\[2)\ \frac{\log_{2}192}{\log_{12}2} - \frac{\log_{2}24}{\log_{96}2} =\]

\[= \log_{2}\left( 2^{3} \cdot 3 \right) \cdot \log_{2}\left( 2^{2} \cdot 3 \right) -\]

\[- \log_{2}\left( 3 \cdot 2^{3} \right) \cdot \log_{2}\left( 2^{5} \cdot 3 \right) =\]

\[= \left( 6 + \log_{2}3 \right)\left( 2 + \log_{2}3 \right) -\]

\[- \left( 3 + \log_{2}3 \right)\left( 5 + \log_{2}3 \right) =\]

\[= 12 + 6\log_{2}3 + 2\log_{2}3 +\]

\[+ \left( \log_{2}3 \right)^{2} - 15 - 3\log_{2}3 -\]

\[- 5\log_{2}3 - \left( \log_{2}3 \right)^{2} = - 3.\]