JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Quadratic Equations and Inequalities

Let $t=e^x$ where $t>0$. The equation becomes:
$t^4+2t^3-t-6=0$

Factoring: $t^3(t+2)-1(t+6)=0$
Trying $t=-2$: $16-16+2-6=-4\ne 0$
Trying $t=1$: $1+2-1-6=-4\ne 0$

By inspection or numerical methods, we can verify that this polynomial has exactly one positive real root. Since $t=e^x>0$, and there is exactly one positive value of $t$ satisfying the equation, there is exactly one real value of $x$.

The answer is 1.