\[\begin{array}{l} {4^x} + {6^x} = {9^x} \Leftrightarrow {\left( {\frac{4}{9}} \right)^x} + {\left( {\frac{6}{9}} \right)^x} = 1 \Leftrightarrow {\left( {\frac{2}{3}} \right)^{2x}} + {\left( {\frac{2}{3}} \right)^x} = 1 \hfill \\ {\left( {\frac{2}{3}} \right)^x} = t \hfill \\ {t^2} + t = 1 \Leftrightarrow \left[ \begin{array}{l} t = \frac{{ - 1 - \sqrt 5 }}{2} \hfill \\ t = \frac{{ - 1 + \sqrt 5 }}{2} \hfill \\ \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} {\left( {\frac{2}{3}} \right)^x} = \frac{{ - 1 - \sqrt 5 }}{2}{\text{ }}\emptyset \hfill \\ {\left( {\frac{2}{3}} \right)^x} = \frac{{ - 1 + \sqrt 5 }}{2} \hfill \\ \end{array} \right. \hfill \\ {\left( {\frac{2}{3}} \right)^x} = \frac{{ - 1 + \sqrt 5 }}{2} \Leftrightarrow x = {\log _{\frac{2}{3}}}\left( {\frac{{\sqrt 5 - 1}}{2}} \right) \hfill \\ \end{array}\]

\[x = {\log _{\frac{2}{3}}}\left( {\frac{{\sqrt 5 - 1}}{2}} \right)\]