Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 294

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\[1)\ \frac{\log_{3}8}{\log_{3}16} = \frac{\log_{3}2^{3}}{\log_{3}2^{4}} =\]

\[= \frac{3 \bullet \log_{3}2}{4 \bullet \log_{3}2} = \frac{3}{4} = 0,75\]

\[2)\ \frac{\log_{5}27}{\log_{5}9} = \frac{\log_{5}3^{3}}{\log_{5}3^{2}} =\]

\[= \frac{3 \bullet \log_{5}3}{2 \bullet \log_{5}3} = \frac{3}{2} = 1,5\]

\[3)\ \frac{\log_{5}36 - \log_{5}12}{\log_{5}9} = \frac{\log_{5}\frac{36}{12}}{\log_{5}3^{2}} =\]

\[= \frac{\log_{5}3}{2 \bullet \log_{5}3} = \frac{1}{2} = 0,5\]

\[4)\ \frac{\log_{7}8}{\log_{7}15 - \log_{7}30} = \frac{\log_{7}8}{\log_{7}\frac{15}{30}} =\]

\[= \frac{\log_{7}8}{\log_{7}\frac{1}{2}} = \frac{\log_{7}2^{3}}{\log_{7}2^{- 1}} =\]

\[= \frac{3 \bullet \log_{7}2}{- 1 \bullet \log_{7}2} = - 3\]