$\underset{{\log }_{e}2}{\overset{x}{\int }}\frac{dt}{\sqrt{e^t-1}}=\frac{\pi }{6}\Rightarrow x=$

$$\begin{gathered} \mathrm{I}={\int }_{{\log }_{\mathrm{e}}^2}^{\mathrm{x}} \frac{\mathrm{dt}}{\sqrt{{\mathrm{e}}^{\mathrm{t}}-1}}=\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$6$}\right. \\ {\mathrm{e}}^{\mathrm{t}}-1={\mathrm{y}}^2 \\ {\mathrm{e}}^{\mathrm{t}} \mathrm{dt}=2\mathrm{y} \mathrm{dy} \\ \mathrm{dt}=\frac{2\mathrm{y} \mathrm{dy}}{1+{\mathrm{y}}^2} \\ \int \frac{1}{\mathrm{y}} \frac{2\mathrm{y} \mathrm{dy}}{1+{\mathrm{y}}^2}=2 {\tan }^{-1} \mathrm{y} \\ =2 {\tan }^{-1} \left(\sqrt{{\mathrm{e}}^{\mathrm{t}}-1}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{I}={\int }_{{\log }_{\mathrm{e}}^2}^{\mathrm{x}} \frac{\mathrm{dt}}{\sqrt{{\mathrm{e}}^{\mathrm{t}}-1}}={\left(2{\tan }^{-1} \sqrt{{\mathrm{e}}^{\mathrm{t}}-1}\right)}_{{\log }_{\mathrm{e}}^2}^{\mathrm{x}}=\frac{\mathrm{\pi }}{6} \\ =2\left[{\tan }^{-1}\sqrt{{\mathrm{e}}^{\mathrm{x}}-1}-{\tan }^{-1} 1\right]=\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$6$}\right. \\ \Rightarrow {\tan }^{-1} \sqrt{{\mathrm{e}}^{\mathrm{x}}-1}-\frac{\mathrm{\pi }}{4}=\frac{\mathrm{\pi }}{12} \\ {\tan }^{-1} \sqrt{{\mathrm{e}}^{\mathrm{x}}-1} =\frac{3\mathrm{\pi }+\mathrm{\pi }}{12}=\frac{\mathrm{\pi }}{3} \\ \sqrt{{\mathrm{e}}^{\mathrm{x}}-1}=\sqrt{3} \\ {\mathrm{e}}^{\mathrm{x}}-1=3 \Rightarrow \mathrm{x}={\log }_{\mathrm{e}}^4 \\ \mathrm{x}=2{\log }_{\mathrm{e}}^2 \end{gathered}$$