This is a tricky problem! Your first step is to remember that ay = b is the same thing as logab = y. For your problem, let a = 2. Then b = 72x and y = x - 6. So you have log2(72x) = x - 6. That may look equally daunting, but here you can make use of the rule that tells you loga(bx) = xlogab. In your case, this means we have 2x(log27) = x - 6.
From this point it becomes a regular algebra problem. How would you solve this problem if it were 2xk = x - 6, where k is some constant? Let me know if you need further help.