Why not test your skills?

$\begin{aligned} \huge \sqrt[5x]{{\left(\sqrt[x^5]{{\left(5\right)}^5}\right)}^{\sqrt[5]{x}}} & = \huge 3125\\ \\ \huge 5^{\tfrac{5}{x^5} × x^{\frac{1}{5}} × \tfrac{1}{5x}} & = \huge 5^5\\ \\ \huge 5^{x^{-5 + \frac{1}{5} - 1}} & = \huge 5^5\\ \\ \huge 5^{x^{-\frac{29}{5}}} & = \huge 5^5\\ \\ \text{Equating the powers}:-\\ \LARGE x^{-\tfrac{29}{5}} & = \LARGE 5\\ \\ \text{Raising power of 5 on both sides}:-\\ \\ \LARGE x^{-29} & = \LARGE 5^5\\ \\ \text{Raising root of -29 on both sides}:-\\ \LARGE x & = \LARGE 5^{-\tfrac{5}{29}}\\ \\ \Large \therefore a + b & = \Large 5 + 29\\ \\ & = \Large \boxed{34} \end{aligned}$

$Let ~ Y=5^5=3125. \ \ So\ the\ problem\ reduces\ to\ \huge \sqrt[5x]{{\left( \sqrt[x^5] {Y} \right)}^{\sqrt[5]{x}}} =Y .\\ \huge \sqrt[5x]{{\left( \sqrt[x^5] {Y} \right)}^{\sqrt[5]{x}}} = \left( \sqrt[x^5] {Y} \right)^{^{\dfrac{x^{\frac 1 5}}{5x} } } \\ =\Large Y^{^{\dfrac 1 {5x^{\frac 4 5} \times x^5} } }=Y^1.\\ Since\ the\ base\ Y\ is\ same\ equating\ powers,\\ \Large \dfrac 1 {5x^{\frac {29} 5} }=1\ \ \ \implies\ 5^{-1} =x^{^{\frac {29} 5}}\\ \Large So\ x=5^{^{ -\frac 5 {29} }} =5^{^{ - \frac a b }} \ \ \therefore\ a+b=\huge \ \ \ \color{#D61F06}{34}$