Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 341

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\[1)\log_{7}(x - 1) \bullet \log_{7}x = \log_{7}x\]

\[\log_{7}x = 0\]

\[\log_{7}x = \log_{7}1\ \]

\[x = 1.\]

\[\log_{7}(x - 1) = 1\]

\[\log_{7}(x - 1) = \log_{7}7\]

\[x - 1 = 7\]

\[x = 8.\]

\[имеет\ смысл\ при:\]

\[x - 1 > 0\ \]

\[x > 1.\]

\[Ответ:\ \ x = 8.\]

\[2)\log_{\frac{1}{3}}x \bullet \log_{\frac{1}{3}}(3x - 2) =\]

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

\[\log_{\frac{1}{3}}(3x - 2) = 0\]

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

\[3x - 2 = 1\]

\[3x = 3\]

\[x = 1.\]

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

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

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

\[имеет\ смысл\ при:\]

\[3x - 2 > 0\]

\[\ x > \frac{2}{3}.\]

\[Ответ:\ \ x = 1.\]

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

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

\[\log_{2}(3x + 1) = 0\]

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

\[3x + 1 = 1\]

\[3x = 0\]

\[x = 0.\]

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

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

\[x = 3^{2}\]

\[x = 9.\]

\[имеет\ смысл\ при:\]

\[3x + 1 > 0\]

\[x > - \frac{1}{3}\text{\ \ \ }и\ \ \ x > 0.\]

\[Ответ:\ \ x = 9.\]

\[4)\log_{\sqrt{3}}(x - 2) \bullet \log_{5}x =\]

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

\[\log_{3^{\frac{1}{2}}}(x - 2) \bullet \log_{5}x =\]

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

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

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

\[\log_{3}(x - 2) = 0\]

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

\[x - 2 = 1\]

\[x = 3.\]

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

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

\[x = 5.\]

\[имеет\ смысл\ при:\]

\[x - 2 > 0\ \]

\[x > 2.\]

\[Ответ:\ \ x_{1} = 3;\ \ x_{2} = 5.\]