\(2{\log _3}\left( {x-2} \right) + {\log _3}{\left( {x-4} \right)^2} = 0.\)
Запишем ОДЗ:
\(\left\{ {\begin{array}{*{20}{c}}{x-2 > 0,\,\,\,\,}\\{{{\left( {x-4} \right)}^2} > 0}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x > 2,}\\{x \ne 4}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x\, \in \,\left( {2;4} \right) \cup \left( {4;\infty } \right).} \right.} \right.\)
\(2{\log _3}\left( {x-2} \right) + {\log _3}{\left( {x-4} \right)^2} = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\log _3}{\left( {x-2} \right)^2} + {\log _3}{\left( {x-4} \right)^2} = 0\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,{\log _3}{\left( {\left( {x-2} \right)\left( {x-4} \right)} \right)^2} = {\log _3}1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\left( {{x^2}-6x + 8} \right)^2} = 1\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{x^2}-6x + 8 = 1,\,}\\{{x^2}-6x + 8 = -1}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{x^2}-6x + 7 = 0,}\\{{x^2}-6x + 9 = 0\,\,}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,} \right.\,} \right.\left[ {\begin{array}{*{20}{c}}{x = 3-\sqrt 2 ,}\\{x = 3 + \sqrt 2 ,}\\{x = 3.\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\)
Корень \(x = 3-\sqrt 2 \) не удовлетворяет ОДЗ.
Ответ: \(3;\,\,\,\,\,3 + \sqrt 2 \).