Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 849

\[1)\ x^{\lg x} = 10\]

\[\log_{x}x^{\lg x} = \log_{x}10\]

\[\lg x = \frac{\lg 10}{\lg x}\]

\[\lg x = \frac{1}{\lg x}\ \ \ \ \ | \bullet \lg x\]

\[\lg^{2}x = 1\]

\[\lg x = \pm 1.\]

\[1)\ \lg x = - 1\]

\[x = 10^{- 1} = 0,1.\]

\[2)\ \lg x = 1\]

\[x = 10.\]

\[Ответ:\ \ 0,1;\ 10.\]

\[2)\ x^{\log_{3}x} = 9x\]

\[\log_{x}x^{\log_{3}x} = \log_{x}{9x}\]

\[\log_{3}x = \log_{x}9 + \log_{x}x\]

\[\log_{3}x = \frac{\log_{3}9}{\log_{3}x} + 1\]

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

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

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

\[y^{2} - y - 2 = 0\]

\[D = 1 + 8 = 9\]

\[y_{1} = \frac{1 - 3}{2} = - 1;\]

\[y_{2} = \frac{1 + 3}{2} = 2.\]

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

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

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

\[x = 3^{2} = 9.\]

\[Ответ:\ \ \frac{1}{3};\ 9.\]

\[3)\ x^{\lg x} - 1 = 10\left( 1 - x^{- \lg x} \right)\]

\[x^{\lg x} - 1 = 10 - 10x^{- \lg x}\]

\[x^{\lg x} + 10x^{- \lg x} - 11 = 0\]

\[y = x^{\lg x}:\]

\[y + 10y^{- 1} - 11 = 0\ \ \ \ \ | \bullet y\]

\[y^{2} - 11y + 10 = 0\]

\[D = 121 - 40 = 81\]

\[y_{1} = \frac{11 - 9}{2} = 1;\text{\ \ }\]

\[y_{2} = \frac{11 + 9}{2} = 10.\]

\[1)\ x^{\lg x} = 1\]

\[\lg x = 0\]

\[x = 10^{0} = 1.\]

\[2)\ x^{\lg x} = 10\]

\[\log_{x}x^{\lg x} = \log_{x}10\]

\[\lg x = \frac{\lg 10}{\lg x}\]

\[\lg x = \frac{1}{\lg x}\ \ \ \ \ | \bullet \lg x\]

\[\lg^{2}x = 1\]

\[\lg x = \pm 1\]

\[\lg x = - 1;\text{\ \ \ }\lg x = 1;\]

\[x_{1} = 10^{- 1} = 0,1;\text{\ \ \ }x_{2} = 10.\]

\[Ответ:\ \ 0,1;\ 1;\ 10.\]

\[4)\ x^{\sqrt{x}} = \sqrt{x^{x}}\]

\[x^{\sqrt{x}} = x^{\frac{x}{2}};\ \ \ x = 1\]

\[\sqrt{x} = \frac{x}{2}\]

\[2\sqrt{x} = x\]

\[4x = x^{2}\]

\[x^{2} - 4x = 0\]

\[x(x - 4) = 0\]

\[x_{1} = 0;\ \text{\ \ }x_{2} = 4.\]

\[Область\ определения:\]

\[x \neq 0;\ \ \ x \geq 0.\]

\[Ответ:\ \ 1;\ 4.\]