#x = log(8)/(log(3/4)), y = (log(8)/(log(3/4))-3)/3#

Explanation:

From #8^y =2^{x-3}# knowing that #8 = 2^3# we can write #2^{3y} = 2^{x-3}# so we conclude that #3y=x-3#. Operating now on #8^{x-y}=3^x->2^{3(x-y)}=3^x# but #3y = x-3# substituting we get #2^{2x+3}=3^x->8 times 4^x= 3^x->(3/4)^x=8#. Finally #x = log(8)/(log(3/4))#. Solving for #y# we get #y = (log(8)/(log(3/4))-3)/3#