Cunning exponents

$\begin{aligned} {\sqrt{x}}^{\sqrt[4]{x}} & = \sqrt[x]{\sqrt[4]{x}}\\ {\left(x^{\tfrac{1}{2}}\right)}^{x^{\frac{1}{4}}} & = {\left(x^{\tfrac{1}{4}}\right)}^{\frac{1}{x}}\\ x^{\tfrac{1}{2}x^{\frac{1}{4}}} & = x^{\tfrac{1}{4x}}\\ \text{Comparing the exponents, we get}:-\\ \dfrac{1}{2}x^{\tfrac{1}{4}} & = \dfrac{1}{4x}\\ x^{\tfrac{1}{4}} & = \dfrac{1}{2x}\\ x^{\tfrac{1}{4}} × x & = \dfrac{1}{2}\\ x^{\tfrac{1}{4} + 1}& = \dfrac{1}{2}\\ x^{\tfrac{5}{4}} & = \dfrac{1}{2}\\ \text{Raising power of 4 on both sides}\\ x^5 & = \dfrac{1}{16}\\ \text{Root of 5 on both sides}\\ x & = \dfrac{1}{\sqrt[5]{16}} \end{aligned}$

$\therefore a = \boxed{16}$

\begin{aligned} \sqrt{x}^\sqrt [4] {x} & = \sqrt [x] {\sqrt [4] {x}} \\ x^{\frac{1}{2}x^\frac{1}{4}} & = \left(x^\frac{1}{4}\right)^\frac{1}{x} \\ \implies \frac{x^\frac{1}{4}}{2} & = \frac{1}{4x} \\ x^{1+\frac{1}{4}} & = \frac{1}{2} \\ x^\frac{5}{4} & = \frac{1}{2} \\ \implies x & = \frac{1}{2^\frac{4}{5}} = \frac{1}{\sqrt [5]{2^4}} = \frac{1}{\sqrt [5]{16}} \end{aligned}

$\implies a = \boxed{16}$