\({2^x} \cdot {3^{\frac{{3x}}{{4\left( {1-x} \right)}}}} = \sqrt[4]{6}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{2^x} \cdot {3^{\frac{{3x}}{{4\left( {1-x} \right)}}}} = {2^{\frac{1}{4}}} \cdot {3^{\frac{1}{4}}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{{3^{\frac{{3x}}{{4\left( {1-x} \right)}}}}}}{{{3^{\frac{1}{4}}}}} = \dfrac{{{2^{\frac{1}{4}}}}}{{{2^x}}}\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,{3^{\frac{{3x}}{{4\left( {1-x} \right)}}-\frac{1}{4}}} = {2^{\frac{1}{4}-x}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{3^{\frac{{4x-1}}{{4\left( {1-x} \right)}}}} = {2^{\frac{{1-4x}}{4}}}.\)
Прологарифмируем обе части последнего уравнения по основанию 3:
\({\log _3}{3^{\frac{{4x-1}}{{4\left( {1-x} \right)}}}} = {\log _3}{2^{\frac{{1-4x}}{4}}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{4x-1}}{{4\left( {1-x} \right)}}{\log _3}3 = \dfrac{{1-4x}}{4}{\log _3}2\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{4x-1}}{{4\left( {1-x} \right)}} + \dfrac{{4x-1}}{4}{\log _3}2 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{4x-1}}{4} \cdot \left( {\dfrac{1}{{1-x}} + {{\log }_3}2} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{4x-1 = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\dfrac{1}{{1-x}} + {{\log }_3}2 = 0}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 0,25,\,\,\,\,\,\,\,\,\,\,\,}\\{1-x = -{{\log }_2}3}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{0,25,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x = 1 + {{\log }_2}3.}\end{array}} \right.} \right.} \right.\)
Ответ: \(\dfrac{1}{4};\,\,\,\,\,1 + {\log _2}3\).