Задача 13 Решите систему уравнений \(\left\{ {\begin{array}{*{20}{c}} {20\, \cdot {x^{{{\log }_3}y}} + 7\, \cdot {y^{{{\log }_3}x}} = 81\sqrt[3]{3}} \\ {x\,y = 9\,\sqrt[3]{9}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \end{array}} \right.\)
Ответ
ОТВЕТ: \(\left( {9;\,\sqrt[3]{9}} \right),\;\;\;\left( {\sqrt[3]{9};\,9} \right).\)
Решение
\(\left\{ {\begin{array}{*{20}{c}}{20 \cdot {x^{{{\log }_3}y}} + 7 \cdot {y^{{{\log }_3}x}} = 81\sqrt[3]{3},}\\{x\,y = 9 \cdot \sqrt[3]{9}.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\)
Запишем ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{x > 0,}\\{y > 0.}\end{array}} \right.\)
Упростим первое уравнение. Для этого воспользуемся свойством: \({a^{{{\log }_c}b}} = {b^{{{\log }_c}a}}.\)
\(20 \cdot {x^{{{\log }_3}y}} + 7 \cdot {y^{{{\log }_3}x}} = 81\sqrt[3]{3}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,20 \cdot {x^{{{\log }_3}y}} + 7 \cdot {x^{{{\log }_3}y}} = 81\sqrt[3]{3}\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,27 \cdot {x^{{{\log }_3}y}} = 81 \cdot \sqrt[3]{3}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{x^{{{\log }_3}y}} = 3 \cdot \sqrt[3]{3}.\)
Прологарифмируем обе части последнего уравнения по основанию 3:
\({x^{{{\log }_3}y}} = 3 \cdot \sqrt[3]{3}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{\log _3}{x^{{{\log }_3}y}} = {\log _3}\left( {3 \cdot \sqrt[3]{3}} \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{\log _3}y \cdot {\log _3}x = \dfrac{4}{3}.\)
Из второго уравнения: \(y = \dfrac{{9 \cdot \sqrt[3]{9}}}{x}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,y = \dfrac{{{3^{\frac{8}{3}}}}}{x}.\)
\({\log _3}y \cdot {\log _3}x = \dfrac{4}{3}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{\log _3}\dfrac{{{3^{\frac{8}{3}}}}}{x} \cdot {\log _3}x = \dfrac{4}{3}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left( {\dfrac{8}{3}-{{\log }_3}x} \right){\log _3}x = \dfrac{4}{3}\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,3\log _3^2x-8{\log _3}x + 4 = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}{\log _3}x = 2,\\{\log _3}x = \dfrac{2}{3}\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = 9,\\x = \sqrt[3]{9}.\end{array} \right.\)
С учётом того, что \(y = \dfrac{{9 \cdot \sqrt[3]{9}}}{x}\), получим: \(\left[ \begin{array}{l}\left\{ \begin{array}{l}x = 9,\\y = \sqrt[3]{9},\end{array} \right.\\\left\{ \begin{array}{l}x = \sqrt[3]{9},\\y = 9.\end{array} \right.\end{array} \right.\)
Ответ: \(\left( {9;\sqrt[3]{9}} \right),\,\,\,\,\left( {\sqrt[3]{9};9} \right).\)