MANHATTAN GMAT OFFICIAL SOLUTION:

As in so many of these problems, the first task is careful reading. Note that x is not itself the integer (except in the case of 1/1); x is the reciprocal of an integer. You might even make a quick table corresponding to the values in the answer choices:

Now, just keep adding columns. Add one for y, defined as –x^2:

Notice that according to PEMDAS, you apply the squaring operation (Exponent) before the negative sign (Subtraction).

Add one more column for x^y. For clarity, we’ll use the caret symbol (^) to indicate exponents. Notice that the negative sign undoes the reciprocal in x, leaving you with an integer base.

Finally, you are left with the task: which of those numbers in the last column is largest? Remember, fractional exponents are roots, so the second number is the fourth root of 2, and so on.

You can quickly eliminate 1, since the positive roots (to any degree) of any integer greater than 1 are always greater than 1. For instance, what is the 25th root of 5? It must be a number bigger than 1, even if only slightly, because that number times itself 25 times in all produces 5. So you can eliminate A.

Let’s compare the second and fourth numbers, because the base of 4 can be rewritten as 2^2. The fourth answer becomes then \((2^2)^{(\frac{1}{16})}=2^{(\frac{1}{8})}\), which is less than 2^(1/4). The number that solves z^8 = 2 is smaller than the one that solves z^4=2. So the answer can’t be D.

What about B versus C? Raise both numbers to the 36th power, so that we eliminate fractional exponents.

2^(1/4)^36=2^9

3^(1/9)^36=3^4

Which of the results is bigger? 2^9=512, while 3^4= only 81. So the original numbers must be in that same order of size; B is bigger than C. C is out.

A similar argument takes out E (raise both B and E to the 100th power):

2^(1/4)^100=2^25

5^(1/25)^100=5^4= 125 = much smaller than 2^25.

The correct answer is B.


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Answer is B. [edited from below]

As corrected below: I originally bit on answer choice E the 25th root of 5 as soon as I knew it was greater than 1. However the 4th root of 2 is greater than the 25th root of 5.

Take the reciprocal of 5 = 1/5 = x

y = -(x^2) therefore y = -(1/25)

x^y = (1/5)^ -(1/25) = 25th square root of 5. To solve negative exponents, take the reciprocal of the base, and apply the exponent as its absolute value. In this case it results: (1/5)^ -(1/25) = 5^(1/25). The 25th square root of 5 is small, but it's greater than 1.

Bunuel, is there any good rule of thumb for comparing the value of square roots?

You're correct. I just checked the math in Excel.

Oversight on my part, I jumped to the conclusion of 5 once I realized it is greater than 1 without out considering the 4th root of 2.

Kudos.

When we input answer choices into the function we end up with the comparison between these values
A)\(1\)
B)\(2^\frac{1}{4}\)
C)\(3^\frac{1}{9}\)
D)\(4^\frac{1}{16} = 2^\frac{1}{8}\)
E)\(5^\frac{1}{25}\)
Obviously A is out, B > D (D is out)
B > C because \(2^\frac{9}{4} = 4*2^\frac{1}{4} > 3\) coz \(2^\frac{1}{4} > 1\) thus C is out
E = \(5^\frac{1}{25} < 8^\frac{1}{25} = 2^\frac{3}{25} < 2^\frac{3}{24} = 2^\frac{1}{8} < 2^\frac{1}{4}\) so E < D < B, thus E is out
B left, B is the answer

I think taking logs helps in these kinds of problems.

The effective equation is --> (1/a^2)*log(a), where a belongs to +I. This is maximum for 2. You since the expression value for a =2 is greater than a= 4.
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GMAT1 650 Q48 V32.