Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 901

\[\boxed{\mathbf{901}.}\]

\[1)\log_{3}x + \log_{9}x + \log_{27}x = \frac{11}{12}\]

\[\log_{3}x + \log_{3^{2}}x + \log_{3^{3}}x = \frac{11}{12}\]

\[\log_{3}x + \frac{1}{2}\log_{3}x + \frac{1}{3}\log_{3}x = \frac{11}{12}\]

\[\log_{3}x \bullet \left( 1 + \frac{1}{2} + \frac{1}{3} \right) = \frac{11}{12}\]

\[\log_{3}x \bullet \left( \frac{6}{6} + \frac{3}{6} + \frac{2}{6} \right) = \frac{11}{12}\]

\[\log_{3}x \bullet \frac{11}{6} = \frac{11}{12}\]

\[\log_{3}x = \frac{1}{2}\]

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

\[x = 3^{\frac{1}{2}}\]

\[x = \sqrt{3}\]

\[Ответ:\ \ x = \sqrt{3}.\]

\[2)\log_{3}x + \log_{\sqrt{3}}x + \log_{\frac{1}{3}}x = 6\]

\[\log_{3}x + \log_{3^{\frac{1}{2}}}x + \log_{3^{- 1}}x = 6\]

\[\log_{3}x + 2\log_{3}x - \log_{3}x = 6\]

\[2\log_{3}x = 6\]

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

\[\log_{3}x = \log_{3}3^{3}\]

\[x = 3^{3}\]

\[x = 27\]

\[Ответ:\ \ x = 27.\]

\[3)\log_{3}x \bullet \log_{2}x = 4\log_{3}2\]

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

\[\left( \log_{3}x \right)^{2} = 4\left( \log_{3}2 \right)^{2}\]

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

\[1)\ \log_{3}x = 2\log_{3}2\]

\[\log_{3}x = \log_{3}2^{2}\]

\[x = 2^{2}\]

\[x = 4.\]

\[2)\ \log_{3}x = - 2\log_{3}2\]

\[\log_{3}x = \log_{3}2^{- 2}\]

\[x = 2^{- 2} = \frac{1}{2^{2}}\]

\[x = \frac{1}{4}.\]

\[Ответ:\ \ x_{1} = 4;\ \ x_{2} = \frac{1}{4}.\]

\[4)\log_{5}x \bullet \log_{3}x = 9\log_{5}3\]

\[\log_{5}x \bullet \frac{\log_{5}x}{\log_{5}3} = 9\log_{5}3\]

\[\left( \log_{5}x \right)^{2} = 9\left( \log_{5}3 \right)^{2}\]

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

\[1)\ \log_{5}x = 3\log_{5}3\]

\[\log_{5}x = \log_{5}3^{3}\]

\[x = 3^{3}\]

\[x = 27.\]

\[2)\ \log_{5}x = - 3\log_{5}3\]

\[\log_{5}x = \log_{5}3^{- 3}\]

\[x = 3^{- 3} = \frac{1}{3^{3}}\]

\[x = \frac{1}{27}.\]

\[Ответ:\ \ x_{1} = 27;\ \ x_{2} = \frac{1}{27}.\]