0 \\ t + \frac{2}{9t} - 1 = 0 \\ 9 {t}^{2} - 9t + 2 = 0 \\ d = 81 - 4 \times 9 \times 2 = 9 \\ t = \frac{9 - 3}{18} = \frac{6}{18} = \frac{1}{3} \\ t = \frac{9 + 3}{18} = \frac{12}{18} = \frac{2}{3} \\ {3}^{x} = \frac{1}{3} \\ x = - 1 \\ {3}^{x} = \frac{2}{3} \\ x = log_{3}( \frac{2}{3} ) = log_{3}(2) - log_{3}(3) = log_{3}(2) - 1" alt=" {3}^{x} + 2 \times {3}^{ - x - 2} = 1 \\ {3}^{x} + 2 \times {3}^{ - x} \times {3}^{ - 2} = 1 \\ {3}^{x} + \frac{2}{9} \times \frac{1}{ {3}^{x} } = 1 \\ {3}^{x} = t \ \: \: \: \: \: \: \: \: t > 0 \\ t + \frac{2}{9t} - 1 = 0 \\ 9 {t}^{2} - 9t + 2 = 0 \\ d = 81 - 4 \times 9 \times 2 = 9 \\ t = \frac{9 - 3}{18} = \frac{6}{18} = \frac{1}{3} \\ t = \frac{9 + 3}{18} = \frac{12}{18} = \frac{2}{3} \\ {3}^{x} = \frac{1}{3} \\ x = - 1 \\ {3}^{x} = \frac{2}{3} \\ x = log_{3}( \frac{2}{3} ) = log_{3}(2) - log_{3}(3) = log_{3}(2) - 1" align="absmiddle" class="latex-formula">