\(27^x ÷ 81^{(x + 2y)} = 9 \\ (27)x = 9 × 81^{(x+2y)} \\ (3^3 )^x =32 \times 3^{4(x + 2y)} \\ =3^{(2 + 4x + 8y)}\\ 3^{3x} = 3^{ (2 + 4x + 8y)}\\ 3x = 2 + 4x + 8y\\ 3x − 4x − 8y = 2 … … … (1)\\ x + 4y = 0 … … … (2)\\ − 4y = 2\\ y = (− 2) ÷ 4 = − ½\\ y = − ½\\ \) Substitute the value of y into equation (2) i.e x + 4y = 0 x + 4( − 1/2) = 0 x − 2 = 0 x = 2 ∴ x = 2,y = − ½) Method II \( 27^x ÷ 31^{(x + 2y) }= 9\\ 3^{3x} × 3^{( − 4x − 8y)} = 32\\ 3^{(3x − 8y)} = 32\\ − x − 8y=2 ……… (1)\\ x + 4y = 0 ……… (2)\\ − 4 = 2\\ y= 2/4 = ½\\ y = ½ \) Substitute the value of y into equation 2 x + 4y=0 x + 4 (− 1) ÷ 2) = 0 x − 2 = 0 x = 2 x = 2, y = ½