given $a,x \in (1,\infty]$then $x$ and $\sqrt[x]{a}$ are different numbers, except for a single value of $x$ which satisfies: $$ x^x = a $$ to solve this equation, therefore, it might help to look at the sequence defined by: $$ x_{n+1} = \frac12\bigg(x_n+e^{\frac{\log a}{x_n}}\bigg) $$ a few trials suggest that this sequence does converge to the required result.
if this suggestion is correct, how does one prove the fact?
Taking logarithm:
$$x \log x= \log a$$
$$x= \exp \left( \frac{\log a}{x} \right)$$
$$2x=x+ \exp \left( \frac{\log a}{x} \right)$$
$$x=\frac{1}{2} \left(x+ \exp \left( \frac{\log a}{x} \right) \right)$$
We have obtained the equation for the fixed point iterations method.
Now we need to consider the convergence conditions.
$$f(x)=\frac{1}{2} \left(x+ \exp \left( \frac{\log a}{x} \right) \right)$$
$$f'(x)=\frac{1}{2} \left(1- \frac{\log a}{x^2} \exp \left( \frac{\log a}{x} \right) \right)$$
For the iterations to converge we need to have:
$$\left|\frac{1}{2} \left(1- \frac{\log a}{x^2} \exp \left( \frac{\log a}{x} \right) \right) \right|<1$$