\({\log _{\sqrt 3 }}x \cdot \sqrt {{{\log }_{\sqrt 3 }}3-{{\log }_x}9} + 4 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,2{\log _3}x \cdot \sqrt {2-2{{\log }_x}3} + 4 = 0\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\sqrt {2-2{{\log }_x}3} = -\dfrac{2}{{{{\log }_3}x}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\sqrt {2-2{{\log }_x}3} = -2{\log _x}3.\)
Пусть \({\log _x}3 = t\). Тогда уравнение примет вид: \(\sqrt {2-2t} = -2t.\)
Уравнение вида: \(\sqrt {f\left( x \right)} = g\left( x \right)\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{f\left( x \right) = {g^2}\left( x \right),}\\{g\left( x \right) \ge 0.\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\)
\(\sqrt {2-2t} = -2t\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{2-2t = 4{t^2},}\\{-2t \ge 0\,\,\,\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{2{t^2} + t-1 = 0}\\{t \le 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\, \Leftrightarrow } \right.} \right.\)
\( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\left[ {\begin{array}{*{20}{c}}{t = -1,}\\{t = \dfrac{1}{2}\,\,\,}\end{array}} \right.}\\{t \le 0\,\,\,\,}\end{array}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,t = -1.} \right.\)
Возвращаясь к прежней переменной, получим:
\({\log _x}3 = -1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{x^{-1}} = 3\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = \dfrac{1}{3}.\)
Ответ: \(\dfrac{1}{3}\).