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