Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 865

\[\boxed{\mathbf{865}.}\]

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

\[\log_{3}(x + 2) < \log_{3}3^{3}\]

\[x + 2 < 3^{3}\]

\[x + 2 < 27\]

\[x < 25.\]

\[имеет\ смысл\ при:\]

\[x + 2 > 0\ \]

\[x > - 2.\]

\[Ответ:\ \ - 2 < x < 25.\]

\[2)\log_{8}(4 - 2x) \geq 2\]

\[\log_{8}(4 - 2x) \geq \log_{8}8^{2}\]

\[4 - 2x \geq 64\]

\[- 2x \geq 60\]

\[x \leq - 30.\]

\[имеет\ смысл\ при:\]

\[4 - 2x > 0\]

\[2x < 4\ \]

\[x < 2.\]

\[Ответ:\ \ x \leq - 30.\]

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

\[\log_{3}(x + 1) < \log_{3}3^{- 2}\]

\[x + 1 < 3^{- 2}\]

\[x < - 1 + \frac{1}{3^{2}}\]

\[x < - 1 + \frac{1}{9}\ \]

\[x < - \frac{8}{9}.\]

\[имеет\ смысл\ при:\]

\[x + 1 > 0\]

\[x > - 1.\]

\[Ответ:\ \ - 1 < x < - \frac{8}{9}.\]

\[4)\log_{\frac{1}{3}}(x - 1) \geq - 2\]

\[\log_{\frac{1}{3}}(x - 1) \geq \log_{\frac{1}{3}}\left( \frac{1}{3} \right)^{- 2}\]

\[x - 1 \leq \left( \frac{1}{3} \right)^{- 2}\]

\[x - 1 \leq 3^{2}\]

\[x \leq 9 + 1\]

\[x \leq 10.\]

\[имеет\ смысл\ при:\]

\[x - 1 > 0\ \]

\[x > 1.\]

\[Ответ:\ \ 1 < x \leq 10.\]

\[5)\log_{\frac{1}{5}}(4 - 3x) \geq - 1\]

\[\log_{\frac{1}{5}}(4 - 3x) \geq \log_{\frac{1}{5}}\left( \frac{1}{5} \right)^{- 1}\]

\[4 - 3x \leq \left( \frac{1}{5} \right)^{- 1}\]

\[4 - 3x \leq 5\]

\[- 3x \leq 1\]

\[x \geq - \frac{1}{3}.\]

\[имеет\ смысл\ при:\]

\[4 - 3x > 0\]

\[- 3x > - 4\]

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

\[Ответ:\ \ - \frac{1}{3} \leq x < 1\frac{1}{3}.\]

\[6)\log_{\frac{2}{3}}(2 - 5x) < - 2\]

\[\log_{\frac{2}{3}}(2 - 5x) < \log_{\frac{2}{3}}\left( \frac{2}{3} \right)^{- 2}\]

\[2 - 5x > \left( \frac{2}{3} \right)^{- 2}\]

\[- 5x > \left( \frac{3}{2} \right)^{2} - 2\]

\[- 5x > \frac{9}{4} - \frac{8}{4}\]

\[- 5x > \frac{1}{4}\]

\[5x < - 0,25\]

\[x < - 0,05.\]

\[имеет\ смысл\ при:\]

\[2 - 5x > 0\]

\[5x < 2\ \]

\[x < 0,4.\]

\[Ответ:\ \ x < - 0,05.\]