\(\sqrt {{{\log }_x}\sqrt {5x} } = -{\log _x}5\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\sqrt {\dfrac{1}{2}{{\log }_x}\left( {5x} \right)} = -{\log _x}5\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\sqrt {\dfrac{1}{2}\left( {{{\log }_x}x + {{\log }_x}5} \right)} = -{\log _x}5\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\sqrt {\dfrac{1}{2}\left( {1 + {{\log }_x}5} \right)} = -{\log _x}5.\)
Пусть \({\log _x}5 = t\). Тогда уравнение примет вид: \(\sqrt {\dfrac{1}{2}\left( {1 + t} \right)} = -t.\)
Получили уравнение вида: \(\sqrt {f\left( x \right)} = g\left( x \right)\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{f\left( x \right) = {g^2}\left( x \right),}\\{g\left( x \right) \ge 0.\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\)
\(\sqrt {\dfrac{1}{2}\left( {1 + t} \right)} = -t\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\dfrac{1}{2}\left( {1 + t} \right) = {t^2},}\\{-t \ge 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\, \Leftrightarrow } \right.\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{2{t^2}-t-1 = 0,}\\{t \le 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\left[ {\begin{array}{*{20}{c}}{t = 1,\,\,\,}\\{t = -\dfrac{1}{2}}\end{array}} \right.}\\{t \le 0\,\,\,\,\,}\end{array}} \right.} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,t = -\dfrac{1}{2}.\)
Возвращаясь к прежней переменной, получим:
\({\log _x}5 = -\dfrac{1}{2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{x^{-\dfrac{1}{2}}} = 5\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = \dfrac{1}{{25}}.\)
Ответ: \(\dfrac{1}{{25}}\).