Question

How to solve :

〖1/(3^(x-2).9^(3x) )=1/(〖√(〖27〗^(2x+3) ))〗^(-1) )〗^

Answer 1

`(1)/(3^(x-2)\cdot9^(3x))=\frac{1}{(\sqrt{27^(2x+3)})^(-1)}\\ (1)/(3^(x-2)\cdot(3^2)^(3x))=\frac{1}{(((3^3)^(2x+3))^{(1)/(2)})^(-1)}\\ (1)/(3^(x-2)\cdot3^(6x))=\frac{1}{((3^(6x+9))^{(1)/(2)})^(-1)}\\ (1)/(3^(x-2+6x))=\frac{1}{(3^{3x+(9)/(2)})^(-1)}\\ (1)/(3^(7x-2))=\frac{1}{3^{-3x-(9)/(2)}}\\ 3^(7x-2)=3^{-3x-(9)/(2)}\\ 7x-2=-3x-(9)/(2)\\ 14x-4=-6x-9\\ 20x=-5\\ x=-4`