Given integral
\(\begin{aligned}
& I=\int_{\log _e 2}^x \frac{d t}{2 \sqrt{e^t-1}}=\frac{\pi}{6} \quad \text{(given)} \\
& \Rightarrow \quad \int_{\log _e 2}^x \frac{e^{-\frac{t}{2}}}{\sqrt{1-\left(e^{-t / 2}\right)^2}} d t=\frac{\pi}{6} \\
& \text {put } e^{-t / 2}=u \text {, at } t=\log _e 2 u=\frac{1}{\sqrt{2}} \text { and at } t=x \text {, } \\
& u=e^{-x / 2} \text { and } e^{-t / 2} d t=-2 d u \\
& \text {so, } I=\int_{\frac{1}{\sqrt{2}}}^{e^{-x / 2}} \frac{-2 d u}{\sqrt{1-u^2}}=\frac{\pi}{6} \\
& \Rightarrow\left[-2 \sin ^{-1} u\right] \int_{\frac{1}{\sqrt{2}}}^{e^{-\frac{x}{2}}}=\frac{\pi}{6} \Rightarrow \sin ^{-1}\left(e^{-x / 2}\right)-\frac{\pi}{4}=-\frac{\pi}{12} \\
& \Rightarrow \sin ^{-1}\left(e^{-x / 2}\right)=\frac{\pi}{6} \Rightarrow e^{-x / 2}=\sin \frac{\pi}{6}=\frac{1}{2} \\
& \Rightarrow-\frac{x}{2}=\log _e\left(\frac{1}{2}\right)=-\log _e(2) \Rightarrow x=2 \cdot \log _e(2) \text {. } \\
\end{aligned}\)
Hence, option (1) is correct.