Probably not answering the questions but this is too long for comments.
Considering the function $$f(x)=a^x-\frac{\log (x)}{\log (a)}$$ its derivatives are
$$f^{(n)}(x)=a^x \log^n(a)+(-1)^n \frac{(n-1)!}{x^n\, \log(a)}$$
The first derivative cancels at two points given by
$$x_1=\frac{W_0\left(\frac{1}{\log (a)}\right)}{\log (a)}\qquad \text{and}\qquad x_2=\frac{W_{-1}\left(\frac{1}{\log (a)}\right)}{\log (a)}$$ which, in the real domain, exist if $\frac{1}{\log (a)}\geq -\frac 1 e$ that is to say if $a \leq e^{-e}$. If this is the case, $f(x_1)<0$ and $f(x_2)>0$ which explains the three roots.
What is interesting is to look at what happens when $a = e^{-e}$. For this value, the solution of $f(x)=0$ is unique $x=\frac 1e$. At this point, the second derivative is also zero and the Taylor expansion is
$$\frac{e^2}{6} \left(x-\frac{1}{e}\right)^3-\frac{5e^3}{24}
\left(x-\frac{1}{e}\right)^4+O\left(\left(x-\frac{1}{e}\right)^5\right)$$ which makes that, at ths point, $x=\frac 1e$ is a triple root of the equation.
On another side, we could also solve the equation for $a$ and its solutions are given by
$$a_1=\left(\frac{x \log (x)}{W_{0}(x \log (x))}\right)^{\frac{1}{x}}\qquad \text{and}\qquad a_2=\left(\frac{x \log (x)}{W_{-1}(x \log (x))}\right)^{\frac{1}{x}}$$ which do exist if $x \leq \frac 1e$. These two functions are worth to be plotted.
When $x \to 1$ the expansion of $a_1$ is
$$a_1=1+(x-1)-(x-1)^2+\frac{1}{2} (x-1)^3+O\left((x-1)^4\right)$$ and using series reversion
$$x= 1+(a_1-1)+O\left((a_1-1)^2\right)$$ making that if $x\to 0 \implies a_1 \to 0$.
$\def\srt{\operatorname{srt}}$
Although there are integral solutions, the super root $\sqrt[n]x_s,\srt_n(x)$ has an expansion using Lagrange reversion:
$$\sqrt[k]x_s=\srt_k(x)= 1+\sum_{n=1}^\infty\frac{\ln^n(x)}{n!}\frac{d^{n-1}\overbrace{e^{t-nte^{te^\cdots}}}^{k-2\ {e^t}\text{s}}}{dt^{n-1}}\bigg|_0$$
Applying general Leibniz rule and $e^z$ Maclaurin expansion each $k-2$ times if $2<k\in\Bbb N$:
$$ \frac{d^{n-1}\overbrace{e^{t-nte^{te^\cdots}}}^{k-2\ {e^t}\text{s}}}{dt^{n-1}}\bigg|_0=\sum_{n_1=0}^\infty\sum_{n_2=0}^{n-1}\dots\sum_{n_{2k-5}=0}^\infty\sum_{n_{2k-4}=0}^{n-1-\sum_\limits{j=1}^{k-3}n_{2j}}\left.\frac{d^{n-1-\sum_\limits{j=1}^{k-2}n_{2j}}}{dt^{n-1-\sum_\limits{j=1}^{k-2}n_{2j}}}e^{(n_{2k-5}+1)t}\right|_0\prod_{j=1}^{k-2}\frac{n_{2j-3}^{n_{2j-1}}}{n_{2j-1}!}\left(\begin{matrix}{n-1-\sum_\limits{m=1}^{j-1}n_{2m}}\\n_{2j}\end{matrix}\right)\left.\frac{d^{n_{2j}}t^{n_{2j-1}}}{dt^{n_{2j}}}\right|_0$$
where $n_{-1}=-n$. Next, use Kronecker delta and factorial power $m^{(n)}$ in $\left.\frac{d^{n_{2j}}t^{n_{2j-1}}}{dt^{n_{2j}}}\right|_0=\delta_{n_{2j},n_{2j-1}}n_{2j-1}^{(n_{2j})}$. As hinted by Quantile Mechanics $(96)$ to $(97)$, remove the odd indexed sums substituting each $n_{2j}=n_{2j-1}$. We reindex $n_{2j}\to n_j$ and simplify:
$$\begin{align}\frac{d^{n-1}\overbrace{e^{t-nte^{te^\cdots}}}^{k-2\ {e^t}\text{s}}}{dt^{n-1}}\bigg|_0=\sum_{n_1=0}^{n-1}\dots \sum_{n_{k-2}=0}^{n-1-\sum\limits_{j=1}^{k-3}n_j} (n_{k-2}+1)^{n-1-\sum\limits_{j=1}^{k-2}n_j}(-1)^{n_1}\prod_{j=1}^{k-2}n_{j-1}^{n_j} \left(\begin{matrix}{n-1-\sum_\limits{m=1}^{j-1}n_m}\\n_j\end{matrix}\right)= \sum_{n_1=0}^{n-1}\dots \sum_{n_{k-1}=0}^{n-1-\sum\limits_{j=1}^{k-2}n_j}(-1)^{n_1}\prod_{j=1}^{k-1}n_{j-1}^{n_j} \left(\begin{matrix}{n-1-\sum_\limits{m=1}^{j-1}n_m}\\n_j\end{matrix}\right)\end{align}$$
where $n_0=n$. These links support the last equality. Therefore:
$$\bbox[1px, border: 2px solid red]{\sqrt[k]z_s=\srt_k(z)=1+\sum_{n=1}^\infty \sum_{n_1=0}^{n-1}\dots\sum_{n_{k-2}=0}^{n-1-\sum\limits_{j=1}^{k-3}n_j}\frac{(-1)^{n_1}\ln^n(z)}{\Gamma \left({n-\sum_\limits{j=1}^{k-2}n_m}\right)(n_{k-2}+1)^{1-n+\sum\limits_{j=1}^{k-2}n_j} n}\prod_{j=1}^{k-2}\frac {n_{j-1}^{n_j}}{n_j!}}$$
where all can be infinite sums or all, except the $n$ sum, can have an upper bound of $n$. Assume improper sums are empty. We find that:
$$\srt_1(z)=z$$
$$\srt_2(z)=1+\ln(z)+\sum_{n=2}^\infty\frac{\ln^n(z)}{n!}(1-n)^{n-1}$$
$$\srt_3(z)=1-\sum_{n=1}^\infty\sum_{k=1}^{n-2}\frac{(-1)^kk^{n-k+1}n^{k-2}\ln^n(z)}{(n-k)!k!}$$
shown here
$$\srt_4(z)=1+\sum_{n=1}^\infty\sum_{k=0}^{n-1}\sum_{m=1}^{n-k-1}\frac{(-1)^k n^{k-1}k^{m-1}\ln^n(z)}{(n-k-m)!k!m!m^{m+k-n-1}}$$
Shown here
$$\srt_5(z)=1+\sum_{n=1}^\infty\sum_{k=0}^{n-1}\sum_{m=0}^{n-k-1}\sum_{j=1}^{n-k-m-1}\frac{(-1)^k k^m n^{k-1} m^{j-1}\ln^n(z)}{(n-j-k-m)!m!k!j!j^{j+m+k-n-1}}$$
Shown here
$$\srt_6(z)=1+\sum_{n=1}^\infty\sum_{k=0}^{n-1}\sum_{m=0}^{n-1-k}\sum_{j=0}^{n-1-k-m}\sum_{p=1}^{n-k-m-j}\frac{(-1)^k n^{k-1}k^m m^j j^{p-1}\ln^n(z)}{(n-1-k-m-j-p)!k!m!j!p!p^{p+j+m+k-n-1}}$$
shown here
$$\vdots$$
$$\srt_\infty(z)=\sqrt[z]z$$
Compare both links for verification. The region of convergence is near $|z|=1$. Help is wanted with the region of convergence for the multiple series expansion and other representations of $\srt_k(z)$.
Best Answer
$\require{begingroup} \begingroup$ $\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$
For $x\ne1$ we can rewrite \eqref{1} as
\begin{align} \frac{\ln(x)}{1-x} &= \frac1k \tag{2}\label{2} . \end{align}
Note that LHS of \eqref{1} is negative for all real $x>0,\ x\ne 1$, and the known unique real solution for $k<0$ is:
\begin{align} x&= \begin{cases} k\Wp\Big(\tfrac1k\,\exp(\tfrac1k)\Big) ,\quad k\in(-1,0) ,\\ k\Wm\Big(\tfrac1k\,\exp(\tfrac1k)\Big) ,\quad k\in(-\infty,-1) \tag{3}\label{3} , \end{cases} \end{align}
where $\Wp$ is the principal branch and $\Wm$ is the other real branch of the Lambert $\W$ function.
$\endgroup$