How to compute the upper (repelling) fixpoint for $b^x=x$ ($1For the computation of the lower (attracting) fixpoint $L_0$ for $b^{L_0} = L_0 $ (where the base $b$ is in the range for convergence $1 \lt b \lt e^{1/e}$) there is the simple formula
$$L_0 = \exp(-W(-\log(b))) \qquad \qquad \text{where $W$ is the Lambert-W-function}$$
or if we directly use the logs for $b$ and $L$ writing $\beta := \log(b)$ and $\lambda := \log(L)$ 
$$\lambda_0 = -W(-\beta) \phantom{\qquad \qquad \text{where $W$ is the Lambert-W-function}}$$
(This has also the nice property that $\beta = \lambda_0 / L_0 $ or $b=L_0^{1/L_0}$.) 
I think there is a related formula for the upper (repelling) fixpoint $L_1$ too, but I can't remember it from the top of my head (some simple variations didn't work so far). Before beginning a long search in older texts and program-sources (the searchkey were much un-sharp) - could someone refresh my memory? A reference would suffice...              
Example: let $b=\sqrt 2$ then $L_0 = 2$ because $b^2 = 2$ and $L_1=4$ because $b^4 = 4$        
Remark: I know how to approximate $L_1$ using the Newton-algorithm - I specifically think to remember that there is some variant with the Lambert W (which is what I want here)
 A: Yes. If you use the other real-valued branch of $W$, denoted $W_{-1}$, you will get this fixed point:
$$\lambda_U = -W_{-1}(-\beta)$$
and
$$L_U = \exp(-W_{-1}(-\log(b)))$$.
But I do not know if or believe that $W_{-1}$ can be expressed through $W = W_0$, the principal branch (unlike simpler multivalued relations like $\sqrt{...}$ and $\log$ in the complex plane.).
A: Another solution which does not involve the LambertW function involves generating the following series inverse, which I will explain in steps.
$$f=\sqrt{2(\exp(x)-x-1)};\;\;\;\;\text{xfixed}=f^{-1}(x)$$
$$\text{xfixed}= x - \frac{x^2}{6}+ \frac{x^3}{36} - \frac{x^4}{270} + \frac{x^5}{4320} + \frac{x^6}{17010} - \frac{139x^7}{5443200} ...$$
In pari-gp, this series would be "serreverse(sqrt((exp(x)-x-1)*2))".  I don't have a closed form for the series.  By plugging both square roots into this series, you can get either fixed point L for the base=b.  
$$L_b = \frac{\text{xfixed}\left(\pm\sqrt{-2(\ln(\ln(b))+1)}\right)-\ln(\ln(b))}{\ln(b)}$$
One interesting thing about the square root radical is that if $b=\exp(1/e)$, then the radical is zero, which makes sense since $b=\exp(1/e)$ has only one fixed point.  Using a 16 term series for xfixed, I got about 19 decimal digits of precision for the two fixed points of $b=\sqrt{2}$  So, why does this work?  We must first show that iterating $$z \mapsto b^z$$ is exactly congruent with iterating 
$$y \mapsto \exp(y)+k;\;\;k=\ln(\ln(b));\;\;\;y=z\ln(b)+\ln(\ln(b))$$
So then our fixed point problem for $z \mapsto b^z$ becomes equivalent to finding the fixed point for $y \mapsto \exp(y)+k$.  If we have the fixed point for y, then we can get the fixed point for $z \mapsto b^z$ by using this equation based on the congruency.
$$z = \frac{y-\ln(ln(b))}{\ln(b)}$$
Next, we do some algebra to get the fixed point for 
$$y = \exp(y)+k;\;\;\;k=\ln\ln(b))$$
$$y = k + 1 + y + \frac{y^2}{2} + \frac{y^3}{6} + \frac{y^4}{24} + \frac{y^5}{5!} ...$$
$$ -(k+1) =  \frac{y^2}{2} + \frac{y^3}{6} + \frac{y^4}{4!} ...$$
$$ -2(k+1) =  y^2 + \frac{2y^3}{6} + \frac{2y^4}{24} + \frac{2y^5}{5!} ...$$
Now we take the square root of both sides.
$$ \sqrt{-2(k+1)} = f(y) = y\cdot \sqrt{1 + \frac{2y}{6} + \frac{2y^2}{24} + \frac{2y^3}{5!} ...}$$
$$ \sqrt{-2(k+1)} = f(y) =
y + \frac{y^2}{6} + \frac{y^3}{36} + \frac{y^4}{270} + \frac{y^5}{2592} + \frac{17y^6}{544320} + \frac{11y^7}{5443200}+...$$
The square root term on the right has a nice defined formal Taylor series in y; which leads exactly to the $f$ in the 2nd line of my post.  So then we can get k from y.  We need the reverse, so we take the inverse of Taylor series for f.
$$ y = f^{-1}\left(\pm\sqrt{-2(k+1)}\right)=f^{-1}\left(\pm\sqrt{-2(\ln(ln(b))+1)}\right)$$
Now we have the fixed point $y=\exp(y)+k$, and the final step is to generate the fixed point $z=b^z$ from y using $z = \frac{y-\ln(ln(b))}{\ln(b)}$
