Solve the equation: $\;$ $\log_3 \left(\log^2_{\frac{1}{2}} x - 3 \log_{\frac{1}{2}} x+ 5\right) = 2$
Given equation: $\;\;$ $\log_3 \left(\log^2_{\frac{1}{2}} x - 3 \log_{\frac{1}{2}} x + 5\right) = 2$
i.e. $\;$ $\log^2_{\frac{1}{2}} x - 3 \log_{\frac{1}{2}} x + 5 = 3^2 = 9$
i.e. $\;$ $\log^2_{\frac{1}{2}} x - 3 \log_{\frac{1}{2}} x - 4 = 0$
i.e. $\;$ $\log^2_{\frac{1}{2}} x - 4 \log_{\frac{1}{2}} x + \log_{\frac{1}{2}} x - 4 = 0$
i.e. $\;$ $\log_{\frac{1}{2}} x \left(\log_{\frac{1}{2}} x - 4\right) + 1 \left(\log_{\frac{1}{2}} x - 4\right) = 0$
i.e. $\;$ $\left(\log_{\frac{1}{2}}x + 1\right) \left(\log_{\frac{1}{2}}x - 4\right) = 0$
i.e. $\;$ $\log_{\frac{1}{2}}x = -1$ $\;$ or $\;$ $\log_{\frac{1}{2}}x = 4$
i.e. $\;$ $x = \left(\dfrac{1}{2}\right)^{-1} = 2$ $\;$ or $\;$ $x = \left(\dfrac{1}{2}\right)^4 = \dfrac{1}{16}$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{1}{16}, \; 2 \right\}$