Let E 1 = { x ∈ ℝ :   x ≠ 1 and x x - 1 > 0 } and E 2 = { x ∈ E 1 :  …

Let $E_1=${$x\in \mathrm{ℝ}: x\ne 1$ and $\frac{x}{x-1}>0$} and $E_2=${$x\in E_1: {\sin }^{-1}\left({\log }_e\left(\frac{x}{x-1}\right)\right)$ is a real number}

(Here the inverse trigonometric function ${\sin }^{-1}\left(x\right)$ assumes values in $\left[-\frac{\mathrm{\pi }}{2}. \frac{\mathrm{\pi }}{2}\right]$.)
Let $f:E_1\to \mathrm{ℝ}$ be the function defined by $f\left(x\right)={\log }_e\left(\frac{x}{x-1}\right)$
And $g:E_2\to \mathbb{R}$ be the function defined by $g\left(x\right)={\sin }^{-1}\left({\log }_e\left(\frac{x}{x-1}\right)\right)$ .