I am interested in the solution $x > 1$ of the transcendental equation $$x^x = ax^b$$ with $a > 1$, $b \in \mathbb{R}$. I am looking for both analytical and numerical methods, as well as approximations.
To see that the solution exists, write the equation as $\log x = A/(x-b)$ with $A = \log a$ and use the elementary bounds of the logarithm to obtain the bounds $$\frac{b+1}{2} + \frac{1}{2} \sqrt{(b-1)^2 + 4A} \leq x \leq \frac{b+1+A}{2} + \frac{1}{2} \sqrt{(b-1+A)^2+4 A}$$ for the solution. This is enough for general purpose numerical methods, but I would like to see if more specialised tools are available.
There is also a potential connection to the Lambert $W$-function: Write the equation as $x\log x = A + b \log x$ and set $y = x \log x$, by which $\log x = W(y)$, so that $$y = A + b \, W(y) \;.$$ This can be iterated with an initial guess based on the above bounds, but perhaps there are other useful results for the $W$-function of which I am not aware.
From a formal point of view, there is probably a solution in terms of the generalized Lambert function (have a look at equation $(4)$).
This being said, consider that you look for the zero's of function $$f(x)=(x-b)\log(x)-\log(a)$$ $$f'(x)=-\frac{b}{x}+\log (x)+1 \qquad \text{and} \qquad f''(x)=\frac{b}{x^2}+\frac{1}{x}$$ The first derivative cancels at $$x_*=\frac{b}{W(e b)}$$ Good news : $f''(x_*) >0 \quad \forall b >0$.
Now, try a Taylor expansion to second order $$f(x)=f(x_*)+\frac 12f''(x_*) (x-x_*)^2+O\left((x-x_*)^3\right)$$ This would give as a first estimate $$x_0=x_*+\sqrt{-2\frac {f(x_*)}{f''(x_*)}}$$
Trying for $a=\pi$ and $b=e$, this would give $x_0=3.26318$ while the solution obtained using Newton method is $x=3.61002$. Using this estimate, the iterates of Newton method are $$\{3.2631771,3.6338427,3.6101047,3.6100159\}$$ Trying for $a=e$ and $b=\pi$, this would give $x_0=3.51825$ while the solution obtained using Newton method is $x=3.87925$.
Try and let me know how this works for you (I have a few ideas for possible improvements).
Edit
From a series of numerical tests I performed, it seems that the left bound you proposed is good for the $x_0$ of Newton method. It can be improved at no cost using $$\log(x) \geq 2\frac{x-1}{x+1}$$ the rhs being the simplest Padé approximant of $\log(x)$ built around $x=1$.
With $A=\log(a)$, this would give $$x_0=\frac{1}{4} \left(A+2b+2+\sqrt{A^2+4 A (b+3)+4 (b-1)^2}\right)$$ which, for the worked examples, gives something quite close to the geometric mean of your present lower and upper bound $(3.71264\text{ vs }3.69213)$ and $(3.97745\text{ vs }3.95909)$.
For $a=\pi$ and $b=e$, Newton iterates are $$\{3.7126445,3.6115922,3.6100163,3.6100159\}$$ For $a=e$ and $b=\pi$, Newton iterates are $$\{3.9774505,3.8806309,3.8792508,3.8792505\}$$
For these two cases, no overshoot during the path to solution.
Another possible approximation could be based on the fact that $$\log(x) > \frac {e \log \left(\frac{e}{2}\right)+\left(x-\frac{e}{2}\right) \left(2+\log \left(\frac{e}{2}\right)\right)} {x+\frac{e}{2} }$$ as soon as $x > 1.36$.
For the worked examples, it could give $3.65464$ and $3.92416$
Update
Since $a >1$, as far as I can see, the solution looked for is $x>b$. So, considering $$(x-b)\log(x)=\log(a)=A$$ approximate $\log(x)$ by its simplest Padé approximant around $x=b$ which is $$\log(x)=\frac {2b\log(b)+ (2+\log (b))(x-b)}{x+b}$$ which leads to $$x_0=\frac{A+4b+\sqrt{A (A+16 b)+4 b \log (b) (A+b \log (b))}}{2 (2+\log (b))}$$
For the worked examples, this gives $x_0=3.61112$ (the solution being $x=3.61002$) and $x_0=3.87962$ (the solution being $x=3.87925$).
Using $a=123456$ and $b=789$ this would give $x_0=790.75687946638$ while the solution is $x= 790.75687946614$.
Using the same idea, we could expand $f(x)+A$ as a series around $x=b$ $$f(x)+A=(x-b)\log(b)+\sum_{n=2}^\infty (-1)^n\frac{ b^{1-n} }{n-1}(x-b)^n$$ and use series reversion to obtain $$x=b+t\left(1+\sum_{n=1}^p c_n u^n\right)$$ where $$L=\log(b) \qquad\qquad t=\frac{A}{L}\qquad\qquad u=-\frac{t}{b L}$$ and the $c_n$'s form the sequence $$\left\{1,\frac{L}{2}+2,\frac{L^2}{3}+\frac{5 L}{2}+5,\frac{L^3}{4}+\frac{11 L^2}{4}+\frac{21 L}{2}+14,\cdots\right\} $$