Quote:
If x>0, x^2=2^64 and x^x=2y then what is the value of y?
I'd recommend pretending there are parentheses around each of the 'x'es in the equation x^x = 2y. That eliminates some of the confusion with how to simplify this. You're allowed to do that, by the way! For instance, if you knew that x = 2+z, then it'd be perfectly valid to write "x^x = (2+z)^(2+z)" with parentheses. You wouldn't have to write it as "2+z^2+z", which is confusing at best.
So, with parentheses, it looks like this:
(2^32)^(2^32) = 2y
In other words,
2^(32*(2^32)) = 2y (because of the rule Bunuel cited)
Look inside of the parentheses first. What's 32*(2^32)? (Ignore the extra 2^ for now - you can add it back in later.) Well, 32 is 2^5, so you've really got (2^5)(2^32) = 2^37.
2^(2^37) = 2y
Here comes the moment of truth - we have to divide the whole thing by 2 in order to find the value of y. That's actually a little tricky. The best approach is to very carefully write out the actual math, and apply the rules even if they seem counterintuitive.
(2^(2^37))/2 = y
2^((2^37)-1) = y
Your answer should look something like that.
Note: I just checked the original problem, and found that it read "2^y" rather than "2y". That's why my answer doesn't agree with the original answer in Bunuel's link!