Solve the equation: $\;$ $\log^2_{\frac{1}{2}} \left(4x\right) + \log_2 \left(\dfrac{x^2}{8}\right) = 8$
Given equation: $\;\;$ $\log^2_{\frac{1}{2}} \left(4x\right) + \log_2 \left(\dfrac{x^2}{8}\right) = 8$
i.e. $\;$ $\dfrac{\log^2_2 \left(4x\right)}{\log^2_2 \left(\frac{1}{2}\right)} + \log_2 x^2 - \log_2 8 = 8$
i.e. $\;$ $\left[\dfrac{\log_2 \left(4x\right)}{\log_2 \left(\frac{1}{2}\right)}\right]^2 + 2 \log_2 x - \log_2 2^3 = 8$
i.e. $\;$ $\left[\dfrac{\log_2 4 + \log_2 x}{\log_2 2^{-1}}\right]^2 + 2 \log_2 x - 3 \log_2 2 = 8$
i.e. $\;$ $\left[\dfrac{\log_2 2^2 + \log_2 x}{-1 \times \log_2 2}\right]^2 + 2 \log_2 x - 3 = 8$
i.e. $\;$ $\left[\dfrac{2 \log_2 2 + \log_2 x}{-1}\right]^2 + 2 \log_2 x - 11 = 0$
i.e. $\;$ $\left[2 + \log_2 x\right]^2 + 2 \log_2 x - 11 = 0$
i.e. $\;$ $4 + \log^2_2 x + 4 \log_2 x + 2 \log_2 x - 11 = 0$
i.e. $\;$ $\log^2_2 x + 6 \log_2 x - 7 = 0$
i.e. $\;$ $\log^2_2 x + 7 \log_2 x - \log_2 x - 7 = 0$
i.e. $\;$ $\log_2 x \left(\log_2 x - 1\right) + 7 \left(\log_2 x - 1\right) = 0$
i.e. $\;$ $\left(\log_2 x + 7\right) \left(\log_2 x - 1\right) = 0$
i.e. $\;$ $\log_2 x = -7$ $\;$ or $\;$ $\log_2 x = 1$
i.e. $\;$ $x = 2^{-7}$ $\;$ or $\;$ $x = 2^1 = 2$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{2^{-7}, \; 2 \right\}$