JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Basic Mathematics

For ${\log }_{(x+1)}(x-2)$ the argument needs $x-2>0$, i.e. $x>2$, and this already makes the base $x+1$ positive and different from $1$; the factor $e^{2\ln x}=x^2$ also needs $x>0$, again secured by $x>2$. The denominator $e^{2\ln x}-(2x+3)=x^2-2x-3=(x-3)(x+1)$ must be non-zero, so $x\ne 3$ (and $x\ne -1$, already excluded). Combining, the function is defined precisely for $x>2$ with $x\ne 3$, that is on $(2,\infty )-\{3\}$.