3^3^3^3^x^3 - Simplifying exponents

The math rule is clear and there is no ambiguity in it.

\((a^m)^n=a^{mn}\)

\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\) (if exponentiation is indicated by stacked symbols, the rule is to work from the top down).

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!

Hey ccooley. Thanks for the awesome explanation! I've also corrected my post above. Thanks!

Original question with solution:

11. If \(x>0\), \(x^2=2^{64}\) and \(x^x=2^y\) then what is the value of \(y\)?
A. 2
B. 2^(11)
C. 2^(32)
D. 2^(37)
E. 2^(64)

\(x^2=2^{64}\) --> \(x=\sqrt{2^{64}}=2^{\frac{64}{2}}=2^{32}\) (note that \(x=-\sqrt{2^{64}}\) is not a valid solution as given that \(x>0\)).

Second step: \(x^x=(2^{32})^{(2^{32})}=2^{32*2^{32}}=2^{2^{5}*2^{32}}=2^{2^{37}}=2^y\) --> \(y=2^{37}\).

OR second step: \(x^x=(2^{32})^x=2^{32x}=2^y\) --> \(y=32x\) --> since \(x=2^{32}\) then \(y=32x=32*2^{32}=2^5*2^{32}=2^{37}\).

Answer: D.