Solve the equation: $\;$ $4^{\frac{2}{x}} - 5 \times 4^{\frac{1}{x}} + 4 = 0$
Given equation: $\;\;$ $4^{\frac{2}{x}} - 5 \times 4^{\frac{1}{x}} + 4 = 0$ $\;\;\; \cdots \; (1)$
Let $\;$ $4^{\frac{1}{x}} = p$ $\;\;\; \cdots \; (2)$
Then $\;\;$ $4^{\frac{2}{x}} = \left[\left(4\right)^{\frac{1}{x}}\right]^2 = p^2$ $\;\;\; \cdots \; (3)$
$\therefore \;$ In view of equations $(2)$ and $(3)$, equation $(1)$ becomes
$p^2 - 5p + 4 = 0$
i.e. $\;$ $\left(p - 4\right) \left(p - 1\right) = 0$
i.e. $\;$ $p = 4$ $\;\;$ or $\;\;$ $p = 1$
When $\;$ $p = 4$, $\;$ we have from equation $(2)$,
$4^{\frac{1}{x}} = 4 = 4^1$ $\implies$ $\dfrac{1}{x} = 1$ $\implies$ $x = 1$
When $\;$ $p = 1$, $\;$ we have from equation $(2)$,
$4^{\frac{1}{x}} = 1 = 4^0$ $\implies$ $\dfrac{1}{x} = 0$ $\implies$ $x = \dfrac{1}{0}$ $\;$ which is not possible.
$\therefore \;$ The solution to the given equation is $\;\;$ $x = 1$.