Solve the equation: $\;$ $\left[\left(2^{\sqrt{x} + 3}\right)^{\frac{1}{2 \sqrt{x}}}\right]^{\frac{2}{\sqrt{x} - 1}} = 2$
Given equation: $\;\;$ $\left[\left(2^{\sqrt{x} + 3}\right)^{\frac{1}{2 \sqrt{x}}}\right]^{\frac{2}{\sqrt{x} - 1}} = 2$
i.e. $\;$ $2^{\frac{\sqrt{x} + 3}{\sqrt{x} \left(\sqrt{x} - 1\right)}} = 2^1$
$\implies$ $\dfrac{\sqrt{x} + 3}{x - \sqrt{x}} = 1$
i.e. $\;$ $\sqrt{x} + 3 = x - \sqrt{x}$
i.e. $\;$ $x - 2 \sqrt{x} - 3 = 0$ $\;\;\; \cdots \; (1)$
Let $\;$ $\sqrt{x} = p$ $\;\;\; \cdots \; (2)$
Then equation $(1)$ becomes
$p^2 - 2p - 3 = 0$
i.e. $\;$ $\left(p - 3\right) \left(p + 1\right) = 0$
i.e. $\;$ $p = 3$ $\;$ or $\;$ $p = -1$
Substituting the value of $p$ in equation $(2)$ gives
when $\;$ $p = 3$, $\;$ $\sqrt{x} = 3$ $\implies$ $x = 9$
when $\;$ $p = -1$, $\;$ $\sqrt{x} = -1$
But square root of a number cannot be negative.
$\implies$ $p = -1$ $\;$ is not a valid solution.
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{9 \right\}$