Just keep logging!

I will tell what I did, tell me if I am wrong,

$(\sqrt{x^2+1} + x) = (2 +\sqrt 3)^{\frac 32}$ and $(\sqrt{x^2+1} -x) = (2 - \sqrt 3)^{\frac 32}$

Squaring on both sides,

${\color{#20A900}{2x^2 + 1}} +2x\sqrt{x^2+1} = {\color{#20A900}{26}} + 15\sqrt 3 \Rightarrow x^2 = \dfrac {25}{2}$

$\Rightarrow x= \dfrac {5}{\sqrt 2} = \dfrac {\sqrt {50}}{2} = \dfrac {\sqrt a}{b}$

Therefore, $a + b +27 = 50 + 2 + 27 = \boxed {79}$

Consider the following:

$\begin{aligned} 2 \pm \sqrt 3 & = \frac {4 \pm 2\sqrt 3}2 = \left(\frac {\sqrt 3 \pm 1}{\sqrt 2} \right)^2 \\ \implies \sqrt{2 \pm \sqrt 3} & = \frac {\sqrt 3 \pm 1}{\sqrt 2} \\ \implies \left(2 \pm \sqrt 3\right)^\frac 32 & = \left(\frac {\sqrt 3 \pm 1}{\sqrt 2}\right)^3 \\ & = \frac {3\sqrt 3 \pm 9 + 3\sqrt 3 \pm 1}{2\sqrt 2} \\ & = \frac {3\sqrt 3 \pm 5}{\sqrt 2} \\ & = \sqrt{\left(\frac 5{\sqrt 2}\right)^2 + 1} \pm \frac 5{\sqrt 2} \end{aligned}$

Therefore, $\log_{2+\sqrt 3} \left(\sqrt{\left(\dfrac 5{\sqrt 2}\right)^2 + 1} + \dfrac 5{\sqrt 2} \right) + \log_{2-\sqrt 3} \left(\sqrt{\left(\dfrac 5{\sqrt 2}\right)^2 + 1} - \dfrac 5{\sqrt 2} \right) = \dfrac 32 + \dfrac 32 = 3$ , $\implies x = \dfrac 5{\sqrt 2} = \dfrac {\sqrt {50}}2$ . And $a+b+27 = 50 + 2 + 27 =\boxed{79}$ .