Ain't it cunning?

$\begin{aligned} \sqrt{{\left(\sqrt[4]{{x}^{5}}\right)}^{3}} & = \sqrt[5]{{\left(\sqrt[3]{{x}^{2x^2}}\right)}^{4}}\\ {\left({\left({\left(x^{5}\right)}^{\tfrac{1}{4}}\right)}^{3}\right)}^{\frac{1}{2}} & = {\left({\left({\left(x^{2x^2}\right)}^{\tfrac{1}{3}}\right)}^{4}\right)}^{\frac{1}{5}}\\ x^{\tfrac{15}{8}} & = x^{\tfrac{8x^2}{15}}\\ \text{Equating the powers}:-\\ \dfrac{15}{8} & = \dfrac{8x^2}{15}\\ x^2 & = \dfrac{225}{8} \end{aligned}$

$\therefore \text{Positive real value of x} = \dfrac{15}{8}\\ = \boxed{1.875}$