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Prelude

This Post is a continuation of this Original Post. The original problem asked is:

How many solutions does the following equation have:

$$
a^x = log_a(x) ,,quad a in (0,1) wedge x inmathbb{R}^+_0
$$

And already has been answered (see Claude Leibovici's answer for details).

Observations

I found this problem interesting and I numerically investigated it a bit deeper. This led me to another question. Figures below show that such roots exist and are reals:

And roots can be computed numerically for several values of $a$:

My observations, so far, are:

1. All roots must lie in $(0,1)$ because $log_a(x) > 0 , forall x in (0,1)$ and $log_a(x) < 0$ elsewhere, and $a^x > 0 ,forall x in mathbb{R}$;


2. Solving this problem involves complex analysis (such as the use of Lambert W function) but the result will stay in the real domain;


3. A "divergence" point occurs at $(e^{-e},e^{-1})$;


4. Solving the original problem is equivalent to solve (base conversion and Lambert W properties):$$ln(a) = frac{Wleft(x ln(x)right)}{x} = frac{ln(x)}{x} Leftrightarrow a_k = expleft[frac{W_kleft(x ln(x)right)}{x}right]$$
   


5. Roots become triple when $a < e^{-e}$ (dashed black vertical line, as shown by Claude Leibovici in his answer)


6. 
   

   Roots have asymptotic behavior, it can be checked in term of $a(x)$ for two branches:

   
   
   
   
   
   • one root tends to unity as $arightarrow 1$: $limlimits_{xrightarrow 1} a = 1$ (green curve rightmost);
   
   
   • two roots tend to zero as $arightarrow 0^+$: $limlimits_{xrightarrow 0^+} a = 0$ (green and orange curves leftmost).
   
   
   
   

Questions:

My main questions are:

• 
  How can I prove that one root tends to unity when base $arightarrow 0^+$ (blue curve leftmost)? By taking the limit of the proper branch.


• 
  Is the point 4 correct? Investigating the solution using Wolfram Alpha it seems both Leibovici and my expressions are equivalent. But the first form suffer a huge float arithmetic error with numpy library. Anyway they can be plotted using the latter form:

Side questions are:

• How is called the point where branches diverge?


• Do end of the branches also have a specific name?


• 
  Can we say that roots are multiple at the "divergence" point? If so, in what sense are they multiple? Claude Leibovici: Roots are multiple in the sense that three first degrees of Taylor expansion vanishes at $x=e^{-1}$ with $a=e^{-e}$.


• Is the green branch a specific one because it behaves smoothly?

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
$$x_1=frac{W_0left(frac{1}{log (a)}right)}{log (a)}qquad text{and}qquad x_2=frac{W_{-1}left(frac{1}{log (a)}right)}{log (a)}$$ 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
$$frac{e^2}{6} left(x-frac{1}{e}right)^3-frac{5e^3}{24}
left(x-frac{1}{e}right)^4+Oleft(left(x-frac{1}{e}right)^5right)$$ 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
$$a_1=1+(x-1)-(x-1)^2+frac{1}{2} (x-1)^3+Oleft((x-1)^4right)$$ and using series reversion
$$x= 1+(a_1-1)+Oleft((a_1-1)^2right)$$ making that if $xto 0 implies a_1 to 0$.