Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 348

\[\boxed{\mathbf{348}\mathbf{.}}\]

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

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

\[\log_{2}x - \frac{2}{\log_{2}x} + 1 = 0\ \ \ \ \ | \bullet \log_{2}x\]

\[\log_{2}^{2}x + \log_{2}x - 2 = 0\]

\[Пусть\ y = \log_{2}x:\]

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

\[D = 1^{2} + 4 \bullet 2 = 1 + 8 = 9\]

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

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

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

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

\[x = 2^{- 2} = \frac{1}{2^{2}} = \frac{1}{4} = 0,25.\]

\[2)\ \log_{2}x = 1\]

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

\[x = 2.\]

\[имеет\ смысл\ при:\]

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

\[Ответ:\ \ x_{1} = 0,25;\ \ x_{2} = 2.\]

\[2)\log_{2}x + \log_{x}2 = 2,5\]

\[\log_{2}x + \frac{\log_{2}2}{\log_{2}x} = 2,5\]

\[2\log_{2}^{2}x - 5\log_{2}x + 2 = 0\]

\[Пусть\ y = \log_{2}x:\]

\[2y^{2} - 5y + 2 = 0\]

\[D = 5^{2} - 4 \bullet 2 \bullet 2 = 25 - 16 = 9\]

\[y_{1} = \frac{5 - 3}{2 \bullet 2} = \frac{2}{4} = \frac{1}{2};\]

\[y_{2} = \frac{5 + 3}{2 \bullet 2} = \frac{8}{4} = 2.\]

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

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

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

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

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

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

\[x = 2^{2}\]

\[x = 4.\]

\[имеет\ смысл\ при:\]

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

\[Ответ:\ \ x_{1} = \sqrt{2};\ \ x_{2} = 4.\]

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

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

\[\log_{3}^{2}x - 3\log_{3}x + 2 = 0\]

\[Пусть\ y = \log_{3}x:\]

\[y^{2} - 3y + 2 = 0\]

\[D = 3^{2} - 4 \bullet 2 = 9 - 8 = 1\]

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

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

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

\[x = 3.\]

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

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

\[x = 3^{2}\]

\[x = 9.\]

\[имеет\ смысл\ при:\]

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

\[Ответ:\ \ x_{1} = 3;\ \ x_{2} = 9.\]

\[4)\log_{3}x - 6\log_{x}3 = 1\]

\[\log_{3}x - 6 \bullet \frac{\log_{3}3}{\log_{3}x} = 1\]

\[\log_{3}x - \frac{6}{\log_{3}x} - 1 = 0\ \ \ \ \ | \bullet \log_{3}x\]

\[\log_{3}^{2}x - \log_{3}x - 6 = 0\]

\[Пусть\ y = \log_{3}x:\]

\[y^{2} - y - 6 = 0\]

\[D = 1^{2} + 4 \bullet 6 = 1 + 24 = 25\]

\[y_{1} = \frac{1 - 5}{2} = - 2;\text{\ \ }\]

\[y_{2} = \frac{1 + 5}{2} = 3.\]

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

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

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

\[x = \frac{1}{9}.\]

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

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

\[x = 3^{3}\]

\[x = 27.\]

\[имеет\ смысл\ при:\]

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

\[Ответ:\ \ x_{1} = \frac{1}{9};\ \ x_{2} = 27.\]