Find x and y if 3^2x-y=1 and 16^x/4=8^3x-y?

Answer 1

you're pretty sloppy with the parentheses, but if you mean
3^(2x-y) = 1
16^(x/4) = 8^(3x-y)
then that is the same as
2^x = 2^(9x-3y)
x = 9x-3y
Now, since 3^0 = 1, that means
2x-y = 0
Now just solve
8x - 3y = 0
2x - y = 0
Looks like (0,0) is the only solution.
Maybe you meant to group the variables in some other way?

Answer 2

I dont know