\({\log _{\sqrt 2 + \sqrt 3 + \sqrt 5 }}x + {\log _x}\left( {\sqrt 2 + \sqrt 3 -\sqrt 5 } \right) = \dfrac{3}{2} + {\log _x}2\sqrt 6 .\)
Запишем ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{x > 0,}\\{x \ne 1}\end{array}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,x\, \in \,\left( {0;1} \right) \cup \left( {1;\infty } \right).} \right.\)
Заметим что:
\(\left( {\sqrt 2 + \sqrt 3 + \sqrt 5 } \right)\left( {\sqrt 2 + \sqrt 3 -\sqrt 5 } \right) = {\left( {\sqrt 2 + \sqrt 3 } \right)^2}-{\left( {\sqrt 5 } \right)^2} = 2 + 2\sqrt 6 + 3-5 = 2\sqrt 6 .\)
\({\log _{\sqrt 2 + \sqrt 3 + \sqrt 5 }}x + {\log _x}\left( {\sqrt 2 + \sqrt 3 -\sqrt 5 } \right) = \dfrac{3}{2} + {\log _x}\left( {\sqrt 2 + \sqrt 3 + \sqrt 5 } \right)\left( {\sqrt 2 + \sqrt 3 -\sqrt 5 } \right)\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,{\log _{\sqrt 2 + \sqrt 3 + \sqrt 5 }}x + {\log _x}\left( {\sqrt 2 + \sqrt 3 -\sqrt 5 } \right) = \dfrac{3}{2} + {\log _x}\left( {\sqrt 2 + \sqrt 3 + \sqrt 5 } \right) + {\log _x}\left( {\sqrt 2 + \sqrt 3 -\sqrt 5 } \right)\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,{\log _{\sqrt 2 + \sqrt 3 + \sqrt 5 }}x-\dfrac{3}{2}-\dfrac{1}{{{{\log }_{\sqrt 2 + \sqrt 3 + \sqrt 5 }}x}} = 0.\)
Пусть \({\log _{\sqrt 2 + \sqrt 3 + \sqrt 5 }}x = t\). Тогда:
\(t-\dfrac{3}{2}-\frac{1}{t} = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,2{t^2}-3t-2 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = 2,\,\,\,\,\,}\\{t = -\dfrac{1}{2}.}\end{array}} \right.\)
Возвращаясь к прежней переменной, получим:
\(\left[ {\begin{array}{*{20}{c}}{{{\log }_{\sqrt 2 + \sqrt 3 + \sqrt 5 }}x = 2,\,\,}\\{{{\log }_{\sqrt 2 + \sqrt 3 + \sqrt 5 }}x = -\dfrac{1}{2}}\end{array}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = {{\left( {\sqrt 2 + \sqrt 3 + \sqrt 5 } \right)}^2},}\\{x = \dfrac{1}{{\sqrt {\sqrt 2 + \sqrt 3 + \sqrt 5 } }}.}\end{array}} \right.} \right.\)
Ответ: \({\left( {\sqrt 2 + \sqrt 3 + \sqrt 5 } \right)^2};\,\,\,\,\,\dfrac{1}{{\sqrt {\sqrt 2 + \sqrt 3 + \sqrt 5 } }}\).