#(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!
