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)?

Solution 1:

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).