Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 906

\[\boxed{\mathbf{906}.}\]

\[1)\log_{\frac{1}{x}}5 + \log_{\frac{1}{x^{2}}}12 +\]

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

\[\log_{x^{- 1}}5 + \log_{x^{- 2}}12 +\]

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

\[- \log_{x}5 - \frac{1}{2}\log_{x}12 +\]

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

\[\log_{x}5^{- 1} - \log_{x}12^{\frac{1}{2}} +\]

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

\[\log_{x}\frac{1}{5} - \log_{x}\sqrt{12} + \log_{x}\sqrt{3} = 1\]

\[\log_{x}\frac{\sqrt{3}}{5\sqrt{12}} = \log_{x}x\]

\[x = \frac{\sqrt{3}}{5\sqrt{12}} = \frac{1}{5\sqrt{4}} = \frac{1}{5 \bullet 2} =\]

\[= \frac{1}{10} = 0,1\]

\[имеет\ смысл\ при:\]

\[x > 0;\ \ x \neq 1.\]

\[Ответ:\ \ x = 0,1.\]

\[2)\ \frac{1}{2}\log_{x}7 - \log_{\frac{1}{\sqrt{x}}}3 -\]

\[- \log_{x^{2}}28 = 1\]

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

\[- \frac{1}{2}\log_{x}28 = 1\]

\[\log_{x}\sqrt{7} + 2\log_{x}3 -\]

\[- \log_{x}28^{\frac{1}{2}} = 1\]

\[\log_{x}\sqrt{7} + \log_{x}3^{2} -\]

\[- \log_{x}\sqrt{28} = 1\]

\[\log_{x}\frac{\sqrt{7} \bullet 3^{2}}{\sqrt{28}} = \log_{x}x\]

\[x = \frac{\sqrt{7} \bullet 3^{2}}{\sqrt{28}} = \frac{9}{\sqrt{4}} = \frac{9}{2} = 4,5\]

\[имеет\ смысл\ при:\]

\[x > 0;\text{\ \ }x \neq 1.\]

\[Ответ:\ \ x = 4,5.\]