Mathematics - Sequence & Series Question with Solution | TestHub

If $1,{\log }_{81}(3^x+48),{\log }_9(3^x-\frac{8}{3})$ are in A.P.,

then $2{\log }_{81}(3^x+48)=1+{\log }_9(3^x-\frac{8}{3})$.

$⟹{\log }_9(3^x+48)=1+{\log }_9(3^x-\frac{8}{3})⟹3^x+48=9(3^x-\frac{8}{3})⟹3^x+48=9\cdot 3^x-24⟹8\cdot 3^x=72⟹3^x=9=3^2$.

Therefore, $x=2$.

Final Answer: The final answer is $\boxed{2}$