\(\left\{ {\begin{array}{*{20}{c}}{{x^{{{\log }_8}y}} + {y^{{{\log }_8}x}} = 4,\,\,}\\{{{\log }_4}x-{{\log }_4}y = 1.}\end{array}} \right.\)
Воспользуемся свойством \({a^{{{\log }_c}b}} = {b^{{{\log }_c}a}}\). Тогда система примет вид:
\(\left\{ {\begin{array}{*{20}{c}}{{x^{{{\log }_8}y}} + {y^{{{\log }_8}x}} = 4,}\\{{{\log }_4}x-{{\log }_4}y = 1}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{{x^{{{\log }_8}y}} + {x^{{{\log }_8}y}} = 4,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{{\log }_4}x = {{\log }_4}y + {{\log }_4}4}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{2 \cdot {x^{{{\log }_8}y}} = 4,\,\,\,\,\,\,\,\,}\\{{{\log }_4}x = {{\log }_4}4y}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \)
\(\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}x > 0,\\y > 0,\\{x^{{{\log }_8}y}} = 2,\\x = 4y\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}x > 0,\\y > 0,\\{\left( {4y} \right)^{{{\log }_8}y}} = 2,\\x = 4y.\end{array} \right.\)
Рассмотрим уравнение \({\left( {4y} \right)^{{{\log }_8}y}} = 2.\) Прологарифмируем обе части по основанию 8:
\({\log _8}{\left( {4y} \right)^{{{\log }_8}y}} = {\log _8}2\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{\log _8}y \cdot \left( {{{\log }_8}4 + {{\log }_8}y} \right) = \dfrac{1}{3}\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\log _8^2y + \frac{2}{3}{\log _8}y-\dfrac{1}{3} = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}{\log _8}y = -1,\\{\log _8}y = \dfrac{1}{3}\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}y = \dfrac{1}{8},\\y = 2.\end{array} \right.\)
Тогда: \(\left[ \begin{array}{l}\left\{ \begin{array}{l}x > 0,\\y > 0,\\y = \dfrac{1}{8},\\x = \dfrac{1}{2},\end{array} \right.\\\left\{ \begin{array}{l}x > 0,\\y > 0,\\y = 2,\\x = 8\end{array} \right.\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}\left\{ \begin{array}{l}x = \dfrac{1}{2},\\y = \dfrac{1}{8},\end{array} \right.\\\left\{ \begin{array}{l}x = 8,\\y = 2.\end{array} \right.\end{array} \right.\)
Ответ: \(\left( {\dfrac{1}{2};\dfrac{1}{8}} \right),\,\,\,\,\,\left( {8;2} \right).\)