Question

Solve the following equation:

${\log }_3\sqrt{130-7^{{\log }_x(6-x)}}=2$

Answer 1

${\log }_3\sqrt{130-7^{{\log }_x(6-x)}}=2⟹130-7^{{\log }_x(6-x)}=81⟹7^{{\log }_x(6-x)}=49⟹{\log }_x(6-x)=2⟹x^2=6-x$

$x^2+x-6=0⟹x=-3,2$