The positive root of the transcendental equation $\ln x-\sqrt{x-1}+1=0$ I numerically solved the transcendental equation
$$\ln x-\sqrt{x-1}+1=0$$
and obtained an approximate value of its positive real root $$x \approx 14.498719188878466465738532142574796767250306535...$$
I wonder if it is possible to express the exact solution in terms of known  mathematical constants and elementary or special functions (I am especially interested in those implemented in Mathematica)?
 A: Yes, it is possible to express this root in terms of special functions implemented in Mathematica.

Start with your equation
$$\ln x-\sqrt{x-1}+1=0,\tag1$$
then take exponents of both sides
$$x\ e^{1-\sqrt{x-1}}=1.\tag2$$
Change the variable
$$z=\sqrt{x-1}-1,\tag3$$
then plug this into $(2)$ and divide both sides by $2$
$$\left(\frac{z^2}2+z+1\right)e^{-z}=\frac12.\tag4$$
Now the left-hand side looks very familiar. Indeed, as it can be seen from DLMF 8.4.8 or the formulae $(2),(3)$ on this MathWorld page, it is a special case (for $a=3$) of the regularized gamma function
$$Q(a,z)=\frac{\Gamma(a,z)}{\Gamma(a)},\tag5$$
implemented in Mathematica as GammaRegularized[a, z]. 
Its inverse with respect to $z$ is denoted as $Q^{-1}(a,s)$ and implemented in Mathematica as InverseGammaRegularized[a, s]. We can use this function to express the positive real root of the equation $(4)$ is a closed form
$$z=Q^{-1}\left(3,\ \frac12\right).\tag6$$
Finally, using $(3)$ we can express the positive real root of your equation $(1)$ as follows:
$$x=\left(Q^{-1}\left(3,\ \frac12\right)+1\right)^2+1.\tag7$$
The corresponding Mathematica expression is  
(InverseGammaRegularized[3, 1/2] + 1)^2 + 1
We can numerically check that substitution of this expression into the left-hand side of the equation $(1)$ indeed yields $0$.
I was not able to express the result in terms of simpler functions (like Lambert W-function).
