ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 907

\[\boxed{\mathbf{907}.}\]

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

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

\[- 2\log_{3}(x + 1) = 0\]

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

\[- 2\log_{3}\ (x + 1) = 0\]

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

\[- 2\log_{3}\ (x + 1) = 0\]

\[\log_{3}\ (x + 1) = t:\]

\[3 + \frac{2}{t} - 2t = 0\ \ \ \ \ \ \ \ | \cdot ( - t)\]

\[2t^{2} - 3t - 2 = 0\]

\[D = 9 + 16 = 25\]

\[t_{1} = \frac{3 + 5}{4} = 2;\ \ \ \ \]

\[t_{2} = \frac{3 - 5}{4} = - \frac{1}{2}.\]

\[Область\ определения:\]

\[x + 1 > 0\]

\[x > - 1.\]

\[\textbf{а)}\ \log_{3}\ (x + 1) = 2\]

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

\[x + 1 = 9\]

\[x = 8.\]

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

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

\[x + 1 = \frac{1}{\sqrt{3}}\]

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

\[x = \frac{\sqrt{3} - 3}{3}.\]

\[Ответ:x = 8;\ \ \ x = \frac{\sqrt{3} - 3}{3}.\]

\[2)\ 1 + 2\log_{x + 2}5 = \log_{5}(x + 2)\]

\[1 + 2\log_{x + 2}5 - \log_{5}(x + 2) = 0\]

\[1 + 2 \cdot \frac{\log_{5}\ 5}{\log_{5}\ (x + 2)} -\]

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

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

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

\[t = \log_{5}\ (x + 2):\]

\[1 + \frac{2}{t} - t = 0\ \ \ \ \ \ \ \ | \cdot ( - t)\]

\[t^{2} - t - 2 = 0\]

\[t_{1} = 2;\ \ \]

\[t_{2} = - 1\ (по\ теореме\ Виета).\]

\[Область\ определения:\]

\[x + 2 > 0\]

\[x > - 2.\]

\[\textbf{а)}\ \log_{5}\ (x + 2) = 2\]

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

\[x + 2 = 25\]

\[x = 23.\]

\[\textbf{б)}\ \log_{5}\ (x + 2) = - 1\]

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

\[x + 2 = \frac{1}{5}\]

\[x + 2 = 0,2\]

\[x = - 1,8.\]

\[Ответ:x = - 1,8;\ \ x = 23.\]

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

\[Область\ определения:\]

\[3 + x > 0 \rightarrow x > - 3;\]

\[1 + x > 0 \rightarrow x > - 1.\]

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

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

\[| \cdot \log_{1 + x}\ (3 + x)\]

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

\[\sqrt{\log_{1 + x}\ (3 + x)^{2}} = \sqrt{1}\]

\[\textbf{а)}\ \log_{1 + x}\ (3 + x) = 1\ \ \ \ \ \ \]

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

\[3 + x = 1 + x\]

\[3 = 1\]

\[нет\ корней.\]

\[\textbf{б)}\ \ \log_{1 + x}\ (3 + x) = - 1.\]

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

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

\[3 + x = \frac{1}{1 + x}\ \ \ \ \ \ \ \ \ \ \ \ | \cdot (1 + x)\]

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

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

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

\[D_{1} = 4 - 2 = 2\]

\[x_{1} = \ - 2 - \sqrt{2}\ (не\ подходит);\ \ \ \]

\[x_{2} = - 2 + \sqrt{2}\ \]

\[Ответ:x = - 2 + \sqrt{2}.\]

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

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

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

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

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

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

\[Пусть\ y = \log_{3x + 7}(5x + 3):\]

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

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

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

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

\[y - 1 = 0\ \]

\[y = 1.\]

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

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

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

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

\[2x = 4\ \]

\[x = 2.\]

\[имеет\ смысл\ при:\]

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

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

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

\[Ответ:\ \ x = 2.\]