Solve the equation: $\;$ $4^{\left(x - \sqrt{x^2 - 5}\right)} - 12 \times 2^{\left(x - 1 - \sqrt{x^2 - 5}\right)} + 8 = 0$
Given equation: $\;\;$ $4^{\left(x - \sqrt{x^2 - 5}\right)} - 12 \times 2^{\left(x - 1 - \sqrt{x^2 - 5}\right)} + 8 = 0$ $\;\;\; \cdots \; (1)$
i.e. $\;$ $\left(2^2\right)^{\left(x - \sqrt{x^2 - 5}\right)} - \dfrac{12}{2} \times 2^{\left(x - \sqrt{x^2 - 5}\right)} + 8 = 0$
i.e. $\;$ $\left[2^{\left(x - \sqrt{x^2 - 5}\right)}\right]^2 - 6 \times 2^{\left(x - \sqrt{x^2 - 5}\right)} + 8 = 0$ $\;\;\; \cdots \; (1a)$
Let $\;$ $2^{\left(x - \sqrt{x^2 - 5}\right)} = p$ $\;\;\; \cdots \; (2)$
Then, equation $(1a)$ becomes,
$p^2 - 6p + 8 = 0$
i.e. $\;$ $\left(p - 4\right) \left(p - 2\right) = 0$
i.e. $\;$ $p = 4$ $\;\;$ or $\;\;$ $p = 2$
Substituting the value of $p$ in equation $(2)$ gives
when $\;$ $p = 4$, $\;$ $2^{\left(x - \sqrt{x^2 - 5}\right)} = 4 = 2^2$
i.e. $\;$ $x - \sqrt{x^2 - 5} = 2$
i.e. $\;$ $\sqrt{x^2 - 5} = x - 2$ $\;\;\; \cdots \; (3)$
Squaring equation $(3)$ gives
$x^2 - 5 = x^2 - 4x + 4$
i.e. $\;$ $4x = 9$ $\implies$ $x = \dfrac{9}{4}$
when $\;$ $p = 2$, $\;$ $2^{\left(x - \sqrt{x^2 - 5}\right)} = 2 = 2^1$
i.e. $\;$ $x - \sqrt{x^2 - 5} = 1$
i.e. $\;$ $\sqrt{x^2 - 5} = x - 1$ $\;\;\; \cdots \; (4)$
Squaring equation $(4)$ gives
$x^2 - 5 = x^2 - 2x + 1$
i.e. $\;$ $2x = 6$ $\implies$ $x = 3$
$\therefore \;$ The solution to the given equation are $\;\;$ $x = \dfrac{9}{4}$, $\;$ $x = 3$