Solve the equation: $\;$ $4^{x + \sqrt{x^2 - 2}} - 5 \times 2^{x - 1 + \sqrt{x^2 - 2}} = 6$
Given equation: $\;\;$ $4^{x + \sqrt{x^2 - 2}} - 5 \times 2^{x - 1 + \sqrt{x^2 - 2}} = 6$
i.e. $\;$ $\left(2^2\right)^{x + \sqrt{x^2 - 2}} - 5 \times 2^{x + \sqrt{x^2 - 2}} \times 2^{-1} = 6$
i.e. $\;$ $\left(2^{x + \sqrt{x - 2}}\right)^2 - \dfrac{5}{2} \times 2^{x + \sqrt{x^2 - 2}} = 6$ $\;\;\; \cdots \; (1)$
Let $\;$ $2^{x + \sqrt{x - 2}} = p$ $\;\;\; \cdots \; (2)$
Then equation $(1)$ becomes
$p^2 - \dfrac{5}{2} p = 6$
i.e. $\;$ $2p^2 - 5p - 12 = 0$
i.e. $\;$ $\left(p - 4\right) \left(2p + 3\right) = 0$
i.e. $\;$ $p = 4$ $\;$ or $\;$ $p = \dfrac{-3}{2}$
Substituting the values of $p$ in equation $(2)$ gives
when $\;$ $p = 4$, $\;$ $2^{x + \sqrt{x - 2}} = 4$
i.e. $\;$ $2^{x + \sqrt{x - 2}} = 2^2$
$\implies$ $x + \sqrt{x - 2} = 2$
i.e. $\;$ $\sqrt{x - 2} = 2 - x$
i.e. $\;$ $x - 2 = \left(2 - x\right)^2$
i.e. $\;$ $x - 2 = 4 - 4x + x^2$
i.e. $\;$ $x^2 - 5x + 6 = 0$
i.e. $\;$ $\left(x - 3\right) \left(x - 2\right) = 0$
i.e. $\;$ $x = 3$ $\;$ or $\;$ $x = 2$
when $\;$ $p = \dfrac{-3}{2}$, $\;$ $2^{x + \sqrt{x - 2}} = \dfrac{-3}{2}$
i.e. $\;$ $x + \sqrt{x - 2} = \log \left(\dfrac{-3}{2}\right)$
But logarithm of a negative number is not defined.
$\therefore \;$ $p = \dfrac{-3}{2}$ $\;$ is not a valid solution.
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{2, \; 3 \right\}$