Solve the equation: $\;$ $\log_x \left(9x^2\right) \times \log^2_3 x = 4$
Given equation: $\;\;$ $\log_x \left(9x^2\right) \times \log^2_3 x = 4$
i.e. $\;$ $\left(\log_x 9 + \log_x x^2\right) \times \log^2_3 x = 4$
i.e. $\;$ $\left(\log_x 3^2 + 2 \log_x x\right) \times \log^2_3 x = 4$
i.e. $\;$ $\left(2 \log_x 3 + 2\right) \times \log^2_3 x = 4$
i.e. $\;$ $\left(\log_x 3 + 1\right) \times \log^2_3 x = 2$
i.e. $\;$ $\log_x 3 \times \log^2_3 x + \log^2_3 x = 2$
i.e. $\;$ $\log^2_3 x + \log_x 3 \times \log_3 x \times \log_3 x - 2 = 0$
i.e. $\;$ $\log^2_3 x + \log_x 3 \times \dfrac{1}{\log_x 3} \times \log_3 x - 2 = 0$
i.e. $\;$ $\log^2_3 x + \log_3 x - 2 = 0$
i.e. $\;$ $\log^2_3 x + 2 \log_3 x - \log_3 x - 2 = 0$
i.e. $\;$ $\log_3 x \left(\log_3 x - 1\right) + 2 \left(\log_3 x - 1\right) = 0$
i.e. $\;$ $\left(\log_3 x + 2\right) \left(\log_3 x - 1\right) = 0$
i.e. $\;$ $\log_3 x = -2$ $\;$ or $\;$ $\log_3 x = 1$
i.e. $\;$ $x = 3^{-2} = \dfrac{1}{3^2} = \dfrac{1}{9}$ $\;$ or $\;$ $x = 3^1 = 3$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{1}{9}, \; 3 \right\}$