Question

32^{2+y} = 2 ^{2y-5}

Answer 1

32^(y+2) = 2^(2y-5)
-32^(y+2) = -1^1·2^(5y+10):
-1^1·2^(2y-5) = -1^1·2^(5y+10)
1 = 1 v y 2y-5 = 5y+10

y = -5