Solve the equation: $\;$ $\log_4 \left(2 \times 4^{x-2} - 1\right) + 4 = 2x$
Given equation: $\;\;$ $\log_4 \left(2 \times 4^{x-2} - 1\right) + 4 = 2x$
i.e. $\;$ $\log_4 \left(2 \times 4^{x-2} - 1\right) = 2x - 4$
i.e. $\;$ $\log_4 \left(2 \times 4^{x-2} - 1\right) = 2\left(x - 2\right)$ $\;\;\; \cdots \; (1)$
Let $\;$ $x - 2 = p$ $\;\;\; \cdots \; (2)$
In view of equation $(2)$, equation $(1)$ becomes
$\log_4 \left(2 \times 4^p - 1\right) = 2p$
i.e. $\;$ $2 \times 4^p - 1 = 4^{2p}$
i.e. $\;$ $\left(4^p\right)^2 - 2 \times 4^p + 1 = 0$
i.e. $\;$ $\left(4^p - 1\right)^2 = 0$
i.e. $\;$ $4^p - 1 = 0$
i.e. $\;$ $4^p = 1 = 4^0$
$\implies$ $p = 0$
Substituting the value of $\;$ $p$ $\;$ in equation $(2)$ gives
$x - 2 = 0$ $\implies$ $x = 2$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{2 \right\}$