\[{4^x} + {6^x} = {9^x}\]

\[{\text{Решите уравнение:}}\] $%{\text{3}} \cdot {\text{1}}{{\text{0}}^x} - 5 \cdot {4^x} + 2 \cdot {25^x} = 0$%

\[2 \cdot {81^{x + 1}} - {36^{x + 1}} - 3 \cdot {16^{x + 1}} = 0\]

\[{\text{Решить уравнение:}}\] $%{\text{6}} \cdot {{\text{4}}^{\frac{1}{{{x^2}}}}} - 13 \cdot {6^{\frac{1}{{{x^2}}}}} + 6 \cdot {9^{\frac{1}{{{x^2}}}}} = 0$%

\[{\text{Решите уравнение:}}\] $%\left| {{2^x} - 1} \right| + \left| {{2^x} - 2} \right| = 1$%

\[4 \cdot {7^{2x + 4}} - {3^{2x + 6}} - 2 \cdot {7^{2x + 3}} + {3^{2x + 3}} = 0\]

\[{\left( {\sqrt {4 - \sqrt {15} } } \right)^x} + {\left( {\sqrt {4 + \sqrt {15} } } \right)^x} = 8\]

\[{\text{Решите уравнение:}}\] $%{27^x} - 4 \cdot {3^{x + 2}} + {3^{5 - x}} = 0.$%

$%\left( {\sqrt 5 - \sqrt 2 } \right){3^x} - \frac{{{3^{4 - x}}}}{{\sqrt 5 + \sqrt 2 }} - \left( {\sqrt 6 - \sqrt 2 } \right){2^{1 - 2x}} + \frac{{{2^{2x - 3}}}}{{\sqrt 6 + \sqrt 2 }} = 0$%

\[\frac{{{{\left( {\sqrt 5 + 2} \right)}^x} - 2\sqrt 5 + {{\left( {\sqrt 5 - 2} \right)}^x}}}{{1 - {{\left( {\sqrt 5 - 2} \right)}^x}}} > 0\]