We have,
$\int_{\log 2}^{x} \frac{1}{\sqrt{e^{y}-1}} d y=\frac{\pi}{6}$
Let $\quad \sqrt{e^{y}-1}=t$
$\Rightarrow \quad e^{y}-1=t^{2}$
$\Rightarrow \quad e^{y}=1+t^{2}$
$\Rightarrow \quad e^{y} d y=2 t d t$
$\Rightarrow \quad\left(1+t^{2}\right) d y=2 t d t$
$\Rightarrow \quad d y=\frac{2 t}{\left(1+t^{2}\right)} d t$
Now, $\int_{1}^{\sqrt{e^{x-1}}} \frac{2 t}{\left(1+t^{2}\right) \times t} d t=\frac{\pi}{6}$
$\begin{array}{lc}\Rightarrow & \int_{1}^{\sqrt{e^{x}-1}} \frac{d t}{1+t^{2}}=\frac{\pi}{12} \\ \Rightarrow & \quad\left[\tan ^{-1} t\right]_{1}^{\sqrt{e^{x}-1}}=\frac{\pi}{12} \\ \Rightarrow \quad & \quad \tan ^{-1} \sqrt{e^{x}-1}-\frac{\pi}{4}=\frac{\pi}{12} \\ \Rightarrow & \quad \tan ^{-1} \sqrt{e^{x}-1}=\frac{\pi}{3} \\ \Rightarrow & \tan \left(\tan ^{-1} \sqrt{e^{x}-1}\right)=\tan \left(\frac{\pi}{3}\right) \\ \Rightarrow & \quad e^{x}-1=3 \Rightarrow e^{x}=4 \\ \Rightarrow & \quad x=\log 4\end{array}$