Solve the equation: $\;$ $\log_{\frac{1}{3}} x - 3 \sqrt{\log_{\frac{1}{3}} x} + 2 = 0$
Given equation: $\;\;$ $\log_{\frac{1}{3}} x - 3 \sqrt{\log_{\frac{1}{3}} x} + 2 = 0$
i.e. $\;$ $\left(\sqrt{\log_{\frac{1}{3}} x}\right)^2 - 3 \sqrt{\log_{\frac{1}{3}} x} + 2 = 0$
i.e. $\;$ $\left(\sqrt{\log_{\frac{1}{3}} x}\right)^2 - 2 \sqrt{\log_{\frac{1}{3}} x} - \sqrt{\log_{\frac{1}{3}} x} + 2 = 0$
i.e. $\;$ $\sqrt{\log_{\frac{1}{3}} x} \left(\sqrt{\log_{\frac{1}{3}} x} - 2\right) - 1 \left(\sqrt{\log_{\frac{1}{3}} x} - 2\right) = 0$
i.e. $\;$ $\left(\sqrt{\log_{\frac{1}{3}} x} - 1\right) \left(\sqrt{\log_{\frac{1}{3}} x} - 2\right) = 0$
i.e. $\;$ $\sqrt{\log_{\frac{1}{3}} x} = 1$ $\;$ or $\;$ $\sqrt{\log_{\frac{1}{3}} x} = 2$
i.e. $\;$ $\log_{\frac{1}{3}} x = 1$ $\;$ or $\;$ $\log_{\frac{1}{3}} x = 4$
$\implies$ $x = \left(\dfrac{1}{3}\right)^1 = \dfrac{1}{3}$ $\;$ or $\;$ $x = \left(\dfrac{1}{3}\right)^4 = \dfrac{1}{81}$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{1}{3}, \; \dfrac{1}{81} \right\}$