C-05-Q056mediumsingle_mcqThe substitution used in solving 32x−2−5⋅3x−2−66=03^{2x-2} - 5 \cdot 3^{x-2} - 66 = 032x−2−5⋅3x−2−66=0 is —aa=3xa = 3^xa=3xba=32xa = 3^{2x}a=3