ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1350

\[\boxed{\mathbf{1350}\mathbf{.}}\]

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

\[y = \log_{2}x:\]

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

\[D = 9 - 8 = 1\]

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

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

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

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

\[x = 2.\]

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

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

\[x = 4.\]

\[Ответ:\ \ x_{1} = 2;\ \ x_{2} = 4.\]

\[2)\ \left( \log_{3}x \right)^{2} + 5 = 2\log_{3}x^{3}\]

\[\left( \log_{3}x \right)^{2} + 5 = 2 \bullet 3\log_{3}x\]

\[\left( \log_{3}x \right)^{2} - 6\log_{3}x + 5 = 0\]

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

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

\[D = 36 - 20 = 16\]

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

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

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

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

\[x = 3.\]

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

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

\[x = 243.\]

\[Ответ:\ \ x_{1} = 3;\ \ x_{2} = 243.\]