Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 403

\[\boxed{\mathbf{403.}}\]

\[1)\log_{2}\left( 2^{x} - 5 \right) - \log_{2}\left( 2^{x} - 2 \right) =\]

\[= 2 - x\]

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

\[\frac{2^{x} - 5}{2^{x} - 2} = 2^{2 - x}\]

\[2^{x} - 5 = 2^{2 - x} \bullet \left( 2^{x} - 2 \right)\]

\[2^{x} - 5 = 2^{2 - x + x} - 2^{2 - x + 1}\]

\[2^{x} - 5 = 2^{2} - 2^{3 - x}\]

\[2^{x} - 5 = 4 - \frac{2^{3}}{2^{x}}\]

\[2^{x} + \frac{8}{2^{x}} - 9 = 0\]

\[Пусть\ y = 2^{x}:\]

\[y + \frac{8}{y} - 9 = 0\ \ \ \ \ | \bullet y\]

\[y^{2} - 9y + 8 = 0\]

\[D = 9^{2} - 4 \bullet 8 = 81 - 32 = 49\]

\[y_{1} = \frac{9 - 7}{2} = 1;\ y_{2} = \frac{9 + 7}{2} = 8.\]

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

\[2^{x} = 2^{0}\ \]

\[x = 0.\]

\[2)\ 2^{x} = 8\]

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

\[x = 3.\]

\[имеет\ смысл\ при:\]

\[1)\ 2^{x} - 5 > 0\]

\[2^{x} > 5\]

\[\log_{2}2^{x} > \log_{2}5\ \]

\[x > \log_{2}5.\]

\[2)\ 2^{x} - 2 > 0\]

\[2^{x} > 2\ \]

\[x > 1.\]

\[Ответ:\ \ x = 3.\]

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

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

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

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

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

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

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

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

\[(3 - x)(1 - x) = 1\]

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

\[x^{2} - 4x + 2 = 0\]

\[D = 4^{2} - 4 \bullet 2 = 16 - 8 = 8\]

\[x = \frac{4 \pm \sqrt{8}}{2} = \frac{4 \pm 2\sqrt{2}}{2} =\]

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

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

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

\[3 - x = 1 - x\]

\[3 = 1\]

\[нет\ корней.\]

\[имеет\ смысл\ при:\]

\[3 - x > 0 \Longrightarrow x < 3;\]

\[1 - x > 0 \Longrightarrow x < 1.\]

\[Ответ:\ \ x = 2 - \sqrt{2}.\]

\[Пусть\ y = \log_{2}\left( 2^{x} + 1 \right):\]

\[y \bullet (1 + y) - 2 = 0\]

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

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

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

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

\[1)\ \log_{2}\left( 2^{x} + 1 \right) = - 2\]

\[\log_{2}\left( 2^{x} + 1 \right) = \log_{2}2^{- 2}\]

\[2^{x} + 1 = 2^{- 2}\]

\[2^{x} + 1 = \frac{1}{4}\]

\[2^{x} = - \frac{3}{4}\]

\[нет\ корней.\]

\[2)\ \log_{2}\left( 2^{x} + 1 \right) = 1\]

\[\log_{2}\left( 2^{x} + 1 \right) = \log_{2}2\]

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

\[2^{x} = 1\]

\[2^{x} = 2^{0}\ \]

\[x = 0\]

\[Ответ:\ \ x = 0.\]

\[4)\log_{3x + 7}(5x + 3) =\]

\[= 2 - \log_{5x + 3}(3x + 7)\]

\[\log_{3x + 7}(5x + 3) =\]

\[= 2 - \frac{\log_{3x + 7}(3x + 7)}{\log_{3x + 7}(5x + 3)}\]

\[\log_{3x + 7}(5x + 3) =\]

\[= 2 - \frac{1}{\log_{3x + 7}(5x + 3)}\]

\[Пусть\ y = \log_{3x + 7}(5x + 3):\]

\[y = 2 - \frac{1}{y}\ \ \ \ \ | \bullet y\]

\[y^{2} = 2y - 1\]

\[y^{2} - 2y + 1 = 0\]

\[(y - 1)^{2} = 0\]

\[y - 1 = 0\ \]

\[y = 1.\]

\[\log_{3x + 7}(5x + 3) = 1\]

\[\log_{3x + 7}(5x + 3) =\]

\[= \log_{3x + 7}(3x + 7)\]

\[5x + 3 = 3x + 7\]

\[2x = 4\ \]

\[x = 2.\]

\[имеет\ смысл\ при:\]

\[5x + 3 > 0 \Longrightarrow x > - 0,6;\]

\[3x + 7 > 0 \Longrightarrow x > - \frac{7}{3};\]

\[3x + 7 \neq 1 \Longrightarrow x \neq - 2.\]

\[Ответ:\ \ x = 2.\]