Решебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 25

\[\boxed{\mathbf{25}\mathbf{.}}\]

\[\textbf{а)}\log_{2}x = \log_{4}{(x + 2)}\]

\[x > 0.\]

\[x + 2 > 0\]

\[x > - 2.\]

\[M = (0; + \infty).\]

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

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

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

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

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

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

\[x_{1} + x_{2} = 1;\ \ x_{1} \cdot x_{2} = - 2\]

\[x_{1} = - 1 < 0\ (не\ подходит);\ \ \]

\[x_{2} = 2.\]

\[Ответ:x = 2.\]

\[\textbf{б)}\log_{9}{(x + 8)} = \log_{3}(x + 2)\]

\[x + 8 > 0\]

\[x > - 8.\]

\[x + 2 > 0\]

\[x > - 2.\]

\[M = ( - 2; + \infty).\]

\[\frac{1}{2}\log_{3}{(x + 8)} = \log_{3}(x + 2)\]

\[\log_{3}(x + 8)^{\frac{1}{2}} = \log_{3}(x + 2)\]

\[\log_{3}\sqrt{x + 8} = \log_{3}(x + 2)\]

\[\sqrt{x + 8} = x + 2\]

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

\[x + 8 = x^{2} + 4x + 4\]

\[x^{2} + 3x - 4 = 0\]

\[x_{1} + x_{2} = - 3;\ \ x_{1} \cdot x_{2} = - 4\]

\[x_{1} = - 4 < - 2\ (не\ подходит);\ \ \]

\[x_{2} = 1.\]

\[Ответ:x = 1.\]

\[\textbf{в)}\log_{25}{(9x + 7)} = \log_{5}{(3 + x)}\]

\[9x + 7 > 0\]

\[9x > - 7\]

\[x > - \frac{7}{9}.\]

\[3 + x > 0\]

\[x > - 3.\]

\[M = \left( - \frac{7}{9}; + \infty \right).\]

\[\frac{1}{2}\log_{5}(9x + 7) = \log_{5}(3 + x)\]

\[\log_{5}(9x + 7)^{\frac{1}{2}} = \log_{5}(3 + x)\]

\[\log_{5}\sqrt{9x + 7} = \log_{5}(3 + x)\]

\[\sqrt{9x + 7} = 3 + x\]

\[9x + 7 = (3 + x)^{2}\]

\[9x + 7 = 9 + 6x + x^{2}\]

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

\[x_{1} + x_{2} = 3;\ \ x_{1} \cdot x_{2} = 2\]

\[x_{1} = 1;\ \ \ x_{2} = 2.\]

\[Ответ:x = 1;x = 2.\]

\[\textbf{г)}\log_{4}{(x + 9)} = \log_{2}{(x - 3)}\]

\[x + 9 > 0\]

\[x > - 9.\]

\[x - 3 > 0\]

\[x > 3.\]

\[M = (3; + \infty).\]

\[\frac{1}{2}\log_{2}{(x + 9)} = \log_{2}{(x - 3)}\]

\[\log_{2}(x + 9)^{\frac{1}{2}} = \log_{2}{(x - 3)}\]

\[\log_{2}\sqrt{x + 9} = \log_{2}{(x - 3)}\]

\[\sqrt{x + 9} = x - 3\]

\[x + 9 = (x - 3)^{3}\]

\[x + 9 = x^{2} - 6x + 9\]

\[x^{2} - 7x = 0\]

\[x(x - 7) = 0\]

\[x = 0 < 3 - не\ подходит.\]

\[x - 7 = 0\]

\[x = 7.\]

\[Ответ:x = 7.\]

\[\textbf{д)}\log_{2}{(x + 3)} - \log_{\frac{1}{2}}(x + 3) = 4\]

\[x + 3 > 0\]

\[x > - 3.\]

\[M = ( - 3; + \infty).\]

\[\log_{2}(x + 3) + \log_{2}(x + 3) = 4\]

\[2\log_{2}(x + 3) = 4\]

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

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

\[\log_{2}{(x + 3)} = \log_{2}4\]

\[x + 3 = 4\]

\[x = 1 > - 3.\]

\[Ответ:x = 1.\]

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

\[x + 2 > 0\]

\[x > - 2.\]

\[M = ( - 2; + \infty).\]

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

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

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

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

\[x + 2 = 3\]

\[x = 1 > - 2.\]

\[Ответ:x = 1.\]