Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1354

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\[1)\ 5^{\log_{3}x^{2}} - 6 \bullet 5^{\log_{3}x} + 5 = 0\]

\[5^{2\log_{3}x} - 6 \bullet 5^{\log_{3}x} + 5 = 0\]

\[y = 5^{\log_{3}x}:\]

\[y^{2} - 6y + 5 = 0\]

\[D = 36 - 20 = 16\]

\[y_{1} = \frac{6 - 4}{2} = 1;\]

\[y_{2} = \frac{6 + 4}{2} = 5.\]

\[1)\ 5^{\log_{3}x} = 1\]

\[\log_{3}x = 0\]

\[\log_{3}x = \log_{3}3^{0}\]

\[x = 1.\]

\[2)\ 5^{\log_{3}x} = 5\]

\[\log_{3}x = 1\]

\[\log_{3}x = \log_{3}3^{1}\]

\[x = 3.\]

\[Ответ:\ \ x_{1} = 1;\ \ x_{2} = 3.\]

\[2)\ 25^{\log_{3}x} - 4 \bullet 5^{\log_{3}x + 1} = 125\]

\[5^{2\log_{3}x} - 4 \bullet 5 \bullet 5^{\log_{3}x} = 125\]

\[y = 5^{\log_{3}x}:\]

\[y^{2} - 4 \bullet 5y = 125\]

\[y^{2} - 20y - 125 = 0\]

\[D = 400 + 500 = 900\]

\[y_{1} = \frac{20 - 30}{2} = - 5;\]

\[y_{2} = \frac{20 + 30}{2} = 25.\]

\[1)\ 5^{\log_{3}x} = - 5\]

\[корней\ нет.\]

\[2)\ 5^{\log_{3}x} = 25\]

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

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

\[x = 9.\]

\[Ответ:\ \ x = 9.\]