Solve the equation: $\;$ $\log_3 \left(\log_9 x + \dfrac{1}{2} + 9^x\right) = 2x$
Given equation: $\;\;$ $\log_3 \left(\log_9 x + \dfrac{1}{2} + 9^x\right) = 2x$
$\implies$ By definition, $\;$ $\log_9 x + \dfrac{1}{2} + 9^x = 3^{2x}$
i.e. $\;$ $\log_9 x + \dfrac{1}{2} + 9^x = \left(3^2\right)^x = 9^x$
i.e. $\;$ $\log_9 x + \dfrac{1}{2} = 0$
i.e. $\;$ $\log_9 x = \dfrac{-1}{2}$
$\implies$ By definition, $\;$ $x = \left(9\right)^{\frac{-1}{2}} = \dfrac{1}{9^{\frac{1}{2}}} = \dfrac{1}{3}$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{1}{3} \right\}$