How can i solve this problem, answer in details please

if 4×2^x×4^y=1 & 9^x×3^y=1/3
Calculate the values of x and y

I think you should get the first equation as power of 2 and the second as power of 3:

4×2^x×4^y=1
2^2 * 2^x * 2^2y = 2^0
2 + x + 2y = 0
x = -2y -2

9^x×3^y=1/3
3^2x * 3^y = 3^(-1)
2x + y = -1

>> 2(-2y-2) + y = -1
-3y = 3
y = -1

x = -2(-1) -2 = 0

4*2^x*4^y=1
2^2 X 2^x X (2^2)^y=1
2^2 X 2^x X 2^2y =1 [Since (a^m)^n = a^mn]
2^(2+x+2y)=1
Since a^0=1,
2+x+2y=0
x+2y=-2========>(1)

9^x X 3^y=1/3
(3^2)^x X 3^y X 3 =1
3^2x X 3^y X 3 =1
3^(2x+y+1)=1
As above,
2x+y+1=0
2x+y=-1
Multiplying both sides by 2
4x+2y=-2===========>(2)

(1)=(2)
x+2y=4x+2y
x-4x=2y-2y
-3x=0
x=0
Substituting x=0 in (2)
4(0) + 2y =-2
2y=-2
y=-1

Therefore the solution is x=0, y=-1.

4*2^x *4^y = 1 => 2^(x+2y) = 2^-2 => x + 2y = -2 ....... (I)

9^x *3^y = 1/3 => 3^(2x+y) = 3^-1 => 2x+y = -1 ....(II)

simple system yielding ans (0, -1)