Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 241

\[\boxed{\mathbf{241}\mathbf{.}}\]

\[1)\ \left\{ \begin{matrix} 4^{x} \bullet 2^{y} = 32 \\ 3^{8x + 1} = 3^{3y} \\ \end{matrix} \right.\ \]

\[1)\ 4^{x} \bullet 2^{y} = 32\]

\[2^{2x} \bullet 2^{y} = 2^{5}\]

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

\[2x + y = 5\]

\[y = 5 - 2x.\]

\[2)\ 3^{8x + 1} = 3^{3(5 - 2x)}\]

\[8x + 1 = 3(5 - 2x)\]

\[8x + 1 = 15 - 6x\]

\[14x = 14\]

\[x = 1.\]

\[y = 5 - 2 \bullet 1 = 5 - 2 = 3.\]

\[Ответ:\ \ (1;\ \ 3).\]

\[2)\ \left\{ \begin{matrix} 3^{3x - 2y} = 81\ \ \\ 3^{6x} \bullet 3^{y} = 27 \\ \end{matrix} \right.\ \]

\[1)\ 3^{6x} \bullet 3^{y} = 27\]

\[3^{6x + y} = 3^{3}\]

\[6x + y = 3\]

\[y = 3 - 6x.\]

\[2)\ 3^{3x - 2(3 - 6x)} = 81\]

\[3^{3x - 6 + 12x} = 3^{4}\]

\[3x - 6 + 12x = 4\]

\[15x = 10\]

\[x = \frac{10}{15} = \frac{2}{3}.\]

\[y = 3 - 6 \bullet \frac{2}{3} = 3 - 4 = - 1.\]

\[Ответ:\ \ \left( \frac{2}{3}; - 1 \right).\]