\[\boxed{\mathbf{844}.}\]
\[1)\log_{7}{(x - 1)}\log_{7}x = \log_{7}x\ \]
\[\log_{7}\ x\left( \log_{7}(x - 1) - 1 \right) = 0\]
\[\textbf{а)}\ \log_{7}\ x = 0\]
\[x = 1.\]
\[\textbf{б)}\ \log_{7}(x - 1) - 1 = 0\]
\[\log_{7}(x - 1) = 1\]
\[x - 1 = 7^{1}\]
\[x - 1 = 7\]
\[x = 8.\]
\[Проверка\ показывает,\ что\ x =\]
\[= 1\ не\ является\ корнем\]
\[\ уравнения.\]
\[Ответ:x = 8.\]
\[2)\log_{\frac{1}{3}}x\log_{\frac{1}{3}}\ (3x - 2) =\]
\[= \log_{\frac{1}{3}}\ (3x - 2)\ \]
\[\log_{\frac{1}{3}}\log_{\frac{1}{3}}(3x - 2) -\]
\[- \log_{\frac{1}{3}}(3x - 2) = 0\]
\[\log_{\frac{1}{3}}(3x - 2)\left( \log_{\frac{1}{3}}x - 1 \right) = 0\]
\[\textbf{а)}\ \log_{\frac{1}{3}}(3x - 2) = 0\]
\[3x - 2 = 1\]
\[3x = 3\]
\[x = 1.\]
\[\textbf{б)}\ \log_{\frac{1}{3}}x - 1 = 0\]
\[\log_{\frac{1}{3}}\ x = 1\]
\[x = \left( \frac{1}{3} \right)^{1}\]
\[x = \frac{1}{3}.\]
\[Проверка\ показывает,\ что\ x =\]
\[= \frac{1}{3}\ не\ является\ корнем\]
\[\ уравнения.\]
\[Ответ:x = 1.\]
\[3)\log_{2}(3x + 1)\log_{3}x =\]
\[= 2\log_{2}{(3x + 1)}\ \]
\[\log_{2}(3x + 1)\log_{3}x -\]
\[- 2\log_{2}(3x + 1) = 0\]
\[\log_{2}(3x + 1)\left( \log_{3}\ x - 2 \right) = 0\]
\[\textbf{а)}\ \log_{2}(3x + 1) = 0\]
\[3x + 1 = 1\]
\[3x = 0\]
\[x = 0.\]
\[\textbf{б)}\ \log_{3}\ x - 2 = 0\]
\[\log_{3}x = 2\]
\[x = 3^{2}\]
\[x = 9.\]
\[Проверка\ показывает,\ что\ x =\]
\[= 0\ не\ является\ корнем\ \]
\[уравнения.\]
\[Ответ:x = 9.\]
\[4)\log_{\sqrt{3}}(x - 2)\log_{5}x =\]
\[= 2\log_{3}{(x - 2)}\]
\[\log_{\sqrt{3}}(x - 2)\log_{5}x -\]
\[- 2\log_{3}(x - 2) = 0\]
\[\log_{\sqrt{3}}(x - 2)\log_{5}x -\]
\[- 2\log_{\left( \sqrt{3} \right)^{2}}(x - 2) = 0\]
\[\log_{\sqrt{3}}(x - 2)\log_{5}x -\]
\[- \log_{\sqrt{3}}(x - 2) = 0\]
\[\log_{\sqrt{3}}(x - 2)\left( \log_{5}\ x - 1 \right) = 0\]
\[\textbf{а)}\ \log_{\sqrt{3}}(x - 2) = 0\]
\[x - 2 = 1\]
\[x = 3.\]
\[\textbf{б)}\ \log_{5}\ x - 1 = 0\]
\[\log_{5}\ x = 1\]
\[x = 5.\]
\[Ответ:x = 3;\ \ x = 5.\]