Special solution to $a+e^a\ln x = x+e^a\ln a = a+e^x\ln a$ I was messing around with the equations of the form $a+e^b\ln c$. I set two variables equal and graphed them and I noticed something that interested me enough to ask about.
Let $x\in\mathbb{R}$. For all $a\in\mathbb{R}$ there exists a real solution to the system
$$a+e^a\ln x = x+e^a\ln a = a+e^x\ln a$$
However there is one real value $a\approx 1.0142286126$ at which we get two solutions, the new one at $x\approx  5.841992948$.
I don't even know if we can solve for the general solution in terms of $a$, but even if we can't I wondered if there could be a technique to find the $a$ value that gives us this second solution regardless.
 A: $x=a$ is clearly a solution.
To find $a$, Consider $$a+e^a\ln x = x+e^a\ln a $$ and solve for $x$. The result is given in terms of Lambert function
$$x_1=-e^a\,W\left(-a e^{-e^{-a} a-a}\right)\tag1$$
Now consider $$x+e^a\ln a = a+e^x\ln a$$ for which the solution is now
$$x_2=-W\left(-a^{-e^a} e^a \log (a)\right)+a-e^a \log (a)\tag2$$
Graphing $f(a)=x_1-x_2$, we see a zero close to $a=1.3$.
Edit
As  Leucippus commented, there is a problem with $x_2$ since the argument of Lambert function must be $\geq - \frac 1 e$ and this happens at $a=1.3098$ which is the solution of $e^a\,\log(a)=1$.
A: For $a+e^a\ln x = x+e^a\ln a = a+e^x\ln a$ there are some oddities. For instance take the first and third segments, ie $a + e^{a} \, \ln x = a + e^{x} \, \ln a$ then $x=a$. For the second and third segments, $x + e^{a} \, \ln a = a + e^{x} \, \ln a$ then $x = a$. For the first and second the solution for $x$ is in terms of the Lambert W function as seen by:
$$a + e^{a} \, \ln x = x + e^{a} \, \ln a$$
has solution
$$x = - a \, e^{a} \, W\left(- e^{-a} \, e^{-e^{-a}} \right).$$
This is the interesting case.
