$\begin{aligned} 32^{x+y} &= 16 ^{x +2y}\\ 2^{5(x+y)} &= 2^{4(x +2y)} \\ \text{then} &. \\ 5x +5y &= 4 x + 8 y \\ x &= 3 y \end{aligned}$

32^(x+y) = 16 ^(x +2y)

2^5(x+y) = 2^4(x+2y)

5(x+y) =4(x+2y)

5x+5y = 4x + 8y

5x-4x = 8y-5y

x = 3y

we can say 32 = 2^5 & 16 = 2^4 so 32 ^ (x+y) = 2^(5 x+5 y) & 16 ^ (x+2 y) = 2 ^ (4 x+8 y) so 2^(5 x+5 y) = 2 ^ (4 x+8 y) so (5 x+5 y) = (4 x+8 y) so x=3 y

32^x+y = 2^5(x+y) & 16^x+2y = 2^4(x+2y) Then,5(x+y) = 4(x+2y) So, 5x+5y = 4x+8y: Then, 5x-4x = 8y - 5y ie, x = 3y

We can solve this equation using the substitution technique and the exponential way.

First, we should find a number that is a common base of 32 and 16:

2^{5(x+y)}=2^{4(x+2y)}

Since both sides have the same base, we can equate the two quantities:

5(x+y)=4(x+2y) 5x+5y=4x+8y 5x-4x+5y=4x+8y-4x x+5y-5y=8y-5y x=\box{3y}

32^(x+y) = 16^(x+2y) ---> 32^x(32^y) = 16^x(16^2y) ---> (32^x)/(16^x) = (16^2y)/(32^y) ---> (32/16)^x = (256^y)/(32^y) ---> 2^x = (256/32)^y ---> 2^x = 8^y ---> 2^x = 2^3y ---> x = 3y

not a big deal just change the whole system in base 2 ... rest will be done easily,,,,,,