Solve the equation: $\;$ $\log_4 \left(\log_2 x\right) + \log_2 \left(\log_4 x\right) = 2$
Given equation: $\;\;$ $\log_4 \left(\log_2 x\right) + \log_2 \left(\log_4 x\right) = 2$ $\;\;\; \cdots \; (1)$
Now, $\;$ $\log_4 x = \dfrac{\log_2 x}{\log_2 4} = \dfrac{\log_2 x}{\log_2 2^2} = \dfrac{\log_2 x}{2 \log_2 2}$
i.e. $\;$ $\log_4 x = \dfrac{\log_2 x}{2}$ $\;\;\; \cdots \; (2)$
In view of equation $(2)$, equation $(1)$ becomes
$\log_4 \left(\log_2 x\right) + \dfrac{1}{2} \log_2 \left(\log_2 x\right) = 2$ $\;\;\; \cdots \; (3)$
Let $\;$ $\log_2 x = p$ $\;\;\; \cdots \; (4)$
Then equation $(3)$ becomes
$\log_4 p + \dfrac{1}{2} \log_2 p = 2$
i.e. $\;$ $2 \log_4 p + \log_2 p = 4$
i.e. $\;$ $2 \times \dfrac{\log_2 p}{\log_2 4} + \log_2 p = 4$
i.e. $\;$ $2 \times \dfrac{\log_2 p}{\log_2 2^2} + \log_2 p = 4$
i.e. $\;$ $2 \times \dfrac{\log_2 p}{2 \log_2 2} + \log_2 p = 4$
i.e. $\;$ $\log_2 p + \log_2 p = 4$
i.e. $\;$ $2 \log_2 p = 4$
i.e. $\;$ $\log_2 p = 2$
i.e. $\;$ $p = 2^2 = 4$
Substituting the value of $p$ in equation $(4)$ gives
$\log_2 x = 4$ $\implies$ $x = 2^4 = 16$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{16 \right\}$