\(27^x ÷ 81^{(x + 2y)} = 9 \\
(27)x = 9 x 81^{(x+2y)} \\
(3^3 )^x =32 \times 3^{4(x + 2y)} \\
=3^{(2 + 4x + 8y)}\\
3^{3x} = 3^{ (2 + 4x + 8y)}\\
3x = 2 + 4x + 8y\\
3x − 4x − 8y = 2 ... ... ... (1)\\
x + 4y = 0 ... ... ... (2)\\
− 4y = 2\\
y = (− 2) ÷ 4 = − ½\\
y = − ½\\ \)
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
\( 27^x ÷ 31^{(x + 2y) }= 9\\
3^{3x} x 3^{( − 4x − 8y)} = 32\\
3^{(3x − 8y)} = 32\\
− x − 8y=2 ......... (1)\\
x + 4y = 0 ......... (2)\\
− 4 = 2\\
y= 2/4 = ½\\
y = ½ \)
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