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

Question

What are the rules regarding higher order exponents?

In principle, it appears that if we have a variable as an exponent of an exponent, we can bring it down and multiply it (e.g. 3^4^(5x+1) = 3^((4)(5x+1)) = 3^(20x+4) = (3^20x)(3^4).

So in theory, can we simplify the following expression as such:
3^(4x+1)^(5y-1)^6z = 3^(4x+1)(5y-1)(6z)?

JeffYin · Answer

Good question! Taking the simple example of 3^3^3, one thing to notice is that you get a different answer if you start evaluating from the left:

(3^3)^3 = 3^9

than if you start evaluating from the right:

3^(3^3) = 3^27

Which is correct? I had to do a little bit of research on this, but my informal internet search suggested that more mathematicians and programmers say that the convention is to start evaluating from the right.

Notice that you get the answer that you did if you evaluate from the left, so I would say that you  can't  simplify the expression the way that you did.

However, I also think that the GMAT would give you parentheses so that there is no ambiguity about how to evaluate nested exponents. I'm curious if you have seen an official question with nested exponents that doesn't have parentheses?

Cheers,
Jeff

bionication · Answer

Hey JeffYin , thanks for the explanation. I was wondering if you had a chance to look further into this topic. The question that threw me off was from Q11 in this problem set: https://gmatclub.com/forum/new-tough-an ... 25956.html If x>0, x^2=2^64 and x^x=2^y then what is the value of y? We know that x=2^32 x^x = 2^32^2^32 Now this could be re-written as 2^(2^5)^(2^5). But why can't this simplify to 2^2^5^2^2^5? And if so, how do we know if we're going from right -> left or left -> right? Bunuel , your help would also be appreciated

Thanks!

Bunuel · Answer

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).

ccooley · Answer

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!

bionication · Answer

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

Bunuel · Answer

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.

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

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