#(2^3)^y = 2^(x - 3)#

#2^(3y) = 2^(x - 3)#

We're now in equal bases, so we can eliminate the bases.

#3y = x - 3#

#3y + 3 = x#

#8^((3y + 3) - y) = 3^(3y + 3)#

#8^(2y + 3) = 3^(3y + 3)#

This can be solved using the property #n^m = a^b -> logn^m = loga^b#

#log8^(2y + 3) = log3^(3y + 3)#

Now, we need to simplify using the property #loga^n = nloga#.

#(2y + 3)log8 = (3y + 3)log3#

#2ylog8 + 3log8 = 3ylog3 + 3log3#

#2ylog8 - 3ylog3 = 3log3 - 3log8#

#y(2log8 - 3log3) = 3log3 - 3log8#

Now, simplifying using the rule #log_a(n) - log_a(m) = log_a(n/m)#:

#y = (log(27/512))/(log(64/27))#

Now, substituting back into the original equation to find x:

#2^(3y) = 2^(x - 3)#

#2^(3( (log(27/512))/(log(64/27)))) = 2^(x - 3)#

#3( (log(27/512))/(log(64/27))) = x - 3#

#(log(19683/134217728))/(log(64/27)) + 3 = x#

The solution set is therefore #{(log(19683/134217728))/(log(64/27)) + 3, (log(27/512))/(log(64/27))} or {-7.23; -3.41}#.

Hopefully this helps!