JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Quadratic Equations and Inequalities

Let $t=(\sqrt{3}+\sqrt{2}{)}^x$. Since $(\sqrt{3}-\sqrt{2})=\frac{1}{\sqrt{3}+\sqrt{2}}$, we have $(\sqrt{3}-\sqrt{2}{)}^x=1/t$, so the equation becomes

Both roots are positive and reciprocals of each other ($(5+2\sqrt{6})(5-2\sqrt{6})=25-24=1$). Since $\sqrt{3}+\sqrt{2}>1$, each value of $t$ gives exactly one real $x={\log }_{\sqrt{3}+\sqrt{2}}t$; because the roots are reciprocals, the two values of $x$ are $x=2$ and $x=-2$ (matching $(\sqrt{3}+\sqrt{2}{)}^2=5+2\sqrt{6}$ and $(\sqrt{3}+\sqrt{2}{)}^{-2}=5-2\sqrt{6}$). So $S=\{2,-2\}$ has 2 elements.