(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!