Решебник по алгебре 11 класс Никольский Параграф 8. Уравнения-следствия Задание 40

\[\boxed{\mathbf{40.}}\]

\[\textbf{а)}\log_{2}(x + 1) = \log_{4}(5x + 1)\]

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

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

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

\[x > - 0,2.\]

\[x + 1 = \sqrt{5x + 1}\]

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

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

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

\[x(x - 3) = 0\]

\[x = 0;\ \ x = 3.\]

\[Проверка.\]

\[x = 0:\]

\[\log_{2}1 = \log_{4}1\]

\[0 = 0.\]

\[x = 3:\]

\[\log_{2}4 = \log_{4}16\]

\[2 = 2.\]

\[Ответ:x = 0;\ \ x = 3.\]

\[\textbf{б)}\log_{3}(x - 2) = \log_{9}(3x - 6)\]

\[\log_{3}(x - 2) = \frac{1}{2}\log_{3}(3x - 6)\]

\[\log_{3}(x - 2) = \log_{3}(3x - 6)^{\frac{1}{2}}\]

\[\log_{3}(x - 2) = \log_{3}\sqrt{3x - 6}\]

\[x > 2.\]

\[x - 2 = \sqrt{3x - 6}\]

\[(x - 2)^{2} = 3x - 6\]

\[x^{2} - 4x + 4 - 3x + 6 = 0\]

\[x^{2} - 7x + 10 = 0\]

\[x_{1} + x_{2} = 7;\ \ x_{1} \cdot x_{2} = 10\]

\[x_{1} = 2\ (не\ корень);\]

\[x_{2} = 5.\]

\[Ответ:x = 5.\]