# 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)?

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Try Lambert W function. –  ᴊ ᴀ s ᴏ ɴ Jun 24 '13 at 6:11
The Inverse Symbolic Calculator, to my surprise, did not recognize it. –  André Nicolas Jun 24 '13 at 6:16
Mathematica couldn't solve it with Solve[] and Reduce[]. –  Fixed Point Jun 24 '13 at 8:21
@ᴊᴀsᴏɴ Have you been able to express it in terms of Lambert W? –  Vladimir Reshetnikov Jun 25 '13 at 21:27
@AndréNicolas Does "to my surpise" mean that you were able to find a rather simple form for the root? –  Vladimir Reshetnikov Jun 25 '13 at 21:29

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).

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A heuristic check confirming that the expression we found is indeed the root of the equation can be done in Mathematica using PossibleZeroQ[With[{x = (InverseGammaRegularized[3, 1/2] + 1)^2 + 1}, Log[x] - Sqrt[x - 1] + 1]] (* True *). –  Vladimir Reshetnikov Jun 25 '13 at 22:19