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МатематикаГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 348
Задание 348
\[\boxed{\mathbf{348}\mathbf{.}}\]
\[1)\log_{2}x - 2\log_{x}2 = - 1\]
\[\log_{2}x - 2 \bullet \frac{\log_{2}2}{\log_{2}x} = - 1\]
\[\log_{2}x - \frac{2}{\log_{2}x} + 1 = 0\ \ \ \ \ | \bullet \log_{2}x\]
\[\log_{2}^{2}x + \log_{2}x - 2 = 0\]
\[Пусть\ y = \log_{2}x:\]
\[y^{2} + y - 2 = 0\]
\[D = 1^{2} + 4 \bullet 2 = 1 + 8 = 9\]
\[y_{1} = \frac{- 1 - 3}{2} = - 2;\ \ \]
\[y_{2} = \frac{- 1 + 3}{2} = 1.\]
\[1)\ \log_{2}x = - 2\]
\[\log_{2}x = \log_{2}2^{- 2}\]
\[x = 2^{- 2} = \frac{1}{2^{2}} = \frac{1}{4} = 0,25.\]
\[2)\ \log_{2}x = 1\]
\[\log_{2}x = \log_{2}2\]
\[x = 2.\]
\[имеет\ смысл\ при:\]
\[x > 0;\ \ \ x \neq 1.\]
\[Ответ:\ \ x_{1} = 0,25;\ \ x_{2} = 2.\]
\[2)\log_{2}x + \log_{x}2 = 2,5\]
\[\log_{2}x + \frac{\log_{2}2}{\log_{2}x} = 2,5\]
\[2\log_{2}^{2}x - 5\log_{2}x + 2 = 0\]
\[Пусть\ y = \log_{2}x:\]
\[2y^{2} - 5y + 2 = 0\]
\[D = 5^{2} - 4 \bullet 2 \bullet 2 = 25 - 16 = 9\]
\[y_{1} = \frac{5 - 3}{2 \bullet 2} = \frac{2}{4} = \frac{1}{2};\]
\[y_{2} = \frac{5 + 3}{2 \bullet 2} = \frac{8}{4} = 2.\]
\[1)\ \log_{2}x = \frac{1}{2}\]
\[\log_{2}x = \log_{2}2^{\frac{1}{2}}\]
\[x = 2^{\frac{1}{2}}\]
\[x = \sqrt{2}.\]
\[2)\ \log_{2}x = 2\]
\[\log_{2}x = \log_{2}2^{2}\]
\[x = 2^{2}\]
\[x = 4.\]
\[имеет\ смысл\ при:\]
\[x > 0;\text{\ \ }x \neq 1.\]
\[Ответ:\ \ x_{1} = \sqrt{2};\ \ x_{2} = 4.\]
\[3)\log_{3}x + 2\log_{x}3 = 3\]
\[\log_{3}x + 2 \bullet \frac{\log_{3}3}{\log_{3}x} = 3\]
\[\log_{3}^{2}x - 3\log_{3}x + 2 = 0\]
\[Пусть\ y = \log_{3}x:\]
\[y^{2} - 3y + 2 = 0\]
\[D = 3^{2} - 4 \bullet 2 = 9 - 8 = 1\]
\[y_{1} = \frac{3 - 1}{2} = 1;\text{\ \ }y_{2} = \frac{3 + 1}{2} = 2.\]
\[1)\ \log_{3}x = 1\]
\[\log_{3}x = \log_{3}3\]
\[x = 3.\]
\[2)\ \log_{3}x = 2\]
\[\log_{3}x = \log_{3}3^{2}\]
\[x = 3^{2}\]
\[x = 9.\]
\[имеет\ смысл\ при:\]
\[x > 0;\ \ x \neq 1.\]
\[Ответ:\ \ x_{1} = 3;\ \ x_{2} = 9.\]
\[4)\log_{3}x - 6\log_{x}3 = 1\]
\[\log_{3}x - 6 \bullet \frac{\log_{3}3}{\log_{3}x} = 1\]
\[\log_{3}x - \frac{6}{\log_{3}x} - 1 = 0\ \ \ \ \ | \bullet \log_{3}x\]
\[\log_{3}^{2}x - \log_{3}x - 6 = 0\]
\[Пусть\ y = \log_{3}x:\]
\[y^{2} - y - 6 = 0\]
\[D = 1^{2} + 4 \bullet 6 = 1 + 24 = 25\]
\[y_{1} = \frac{1 - 5}{2} = - 2;\text{\ \ }\]
\[y_{2} = \frac{1 + 5}{2} = 3.\]
\[1)\ \log_{3}x = - 2\]
\[\log_{3}x = \log_{3}3^{- 2}\]
\[x = 3^{- 2} = \frac{1}{3^{2}}\]
\[x = \frac{1}{9}.\]
\[2)\ \log_{3}x = 3\]
\[\log_{3}x = \log_{3}3^{3}\]
\[x = 3^{3}\]
\[x = 27.\]
\[имеет\ смысл\ при:\]
\[x > 0;\text{\ \ }x \neq 1.\]
\[Ответ:\ \ x_{1} = \frac{1}{9};\ \ x_{2} = 27.\]
