Let $$\int_{\alpha}^{\log_e 4} \frac{dx}{\sqrt{e^x - 1}} = \frac{\pi}{6}$$. Then $$e^{\alpha}$$ and $$e^{-\alpha}$$ are the roots of the equation :

We need to evaluate the definite integral $$\int_{\alpha}^{\log_e 4} \frac{dx}{\sqrt{e^x - 1}} = \frac{\pi}{6}$$.

We introduce the substitution $$e^x - 1 = t^2$$, so that $$e^x = 1 + t^2$$ and $$e^x\,dx = 2t\,dt$$, giving $$dx = \frac{2t\,dt}{1 + t^2}$$.

Substituting into the integral yields $$\int \frac{1}{t}\,\frac{2t\,dt}{1 + t^2} = \int \frac{2\,dt}{1 + t^2} = 2\arctan(t) + C = 2\arctan(\sqrt{e^x - 1}) + C$$.

Next, evaluating the definite integral gives $$\Bigl[2\arctan(\sqrt{e^x - 1})\Bigr]_{\alpha}^{\log_e 4} = \frac{\pi}{6}$$.

At $$x = \log_e 4$$, we have $$\sqrt{e^{\ln 4} - 1} = \sqrt{3}$$, so $$2\arctan(\sqrt{3}) = 2 \cdot \frac{\pi}{3} = \frac{2\pi}{3}$$.

Therefore, $$\frac{2\pi}{3} - 2\arctan(\sqrt{e^{\alpha} - 1}) = \frac{\pi}{6}$$.

Rearranging gives $$2\arctan(\sqrt{e^{\alpha} - 1}) = \frac{2\pi}{3} - \frac{\pi}{6} = \frac{\pi}{2}$$, so $$\arctan(\sqrt{e^{\alpha} - 1}) = \frac{\pi}{4}$$.

This implies $$\sqrt{e^{\alpha} - 1} = 1 \implies e^{\alpha} - 1 = 1 \implies e^{\alpha} = 2$$.

Hence, the two roots of the equation are $$e^{\alpha} = 2$$ and $$e^{-\alpha} = \frac{1}{2}$$.

The sum of the roots is $$2 + \frac{1}{2} = \frac{5}{2}$$ and their product is $$2 \times \frac{1}{2} = 1$$.

Forming the quadratic equation with these roots gives $$x^2 - \frac{5}{2}x + 1 = 0$$, which multiplies through by 2 to yield $$2x^2 - 5x + 2 = 0$$.

Therefore, the required equation is $$2x^2 - 5x + 2 = 0$$.