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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?
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Answer is E.

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 and it's also greater than all other answer choices.


Thanks for reprimand. But in this case I think the correct answer is B because 4th root from 2 will be bigger than 25th root from 5.
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ak1802
Answer is E.

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 and it's also greater than all other answer choices.


Thanks for reprimand. But in this case I think the correct answer is B because 4th root from 2 will be bigger than 25th root from 5.

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.

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x is the reciprocal of a positive integer
x = 1/n
and y = –x^2; y = -(1/n)^2
x^y = \((\frac{1}{n})^{\frac{–1}{n^2}}\)
= \(n^{\frac{1}{n^2}}\)

then the answer choices become
1 = \(1^{\frac{1}{1^2}}\) = \(1^{(\frac{1}{1})}\) = \(1\); smaller than all other numbers
2 = \(2^{\frac{1}{2^2}}\) = \(2^{(\frac{1}{4}))}\) = \(1024^{(\frac{1}{40})}\)
3 = \(3^{\frac{1}{3^2}}\) = \(3^{(\frac{1}{9})}\) = \(243^{(\frac{1}{45})}\)
4 = \(4^{\frac{1}{4^2}}\) = \(4^{(\frac{1}{16})}\) = \(512^{(\frac{1}{72})}\)
5 = \(n^{\frac{1}{5^2}}\) = \(5^{(\frac{1}{25})}\) = \(125^{(\frac{1}{75})}\)

maximum value of x^y is \(2\)

Answer B
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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
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Bunuel
If x is the reciprocal of a positive integer, then the maximum value of x^y, where y = –x^2, is achieved when x is the reciprocal of

A. 1
B. 2
C. 3
D. 4
E. 5

\(x=\frac{1}{n}\), and we have \((\frac{1}{n})^y\).
\(y=-x^2\), => \(x^y\) =\(\sqrt[n^2]{n}\)

We can plug in different values and compare them. It will definitely take some time. But do we need to do this. The answer is no.

If we look carefully at this expression we’ll notice that \(n^2\) is growing much faster than \(n\). Saying this in other words, the bigger the value of \(n\), the smaller the value of the expression \(x^y\).

In fact we can take \(2\) from the start. \(n=1\) is the minimum value.

By noticing that \(\sqrt[n^2]{n}\) will always be bigger than 1 for n>1, we got only suitable choice for max value of \(x^y\) as \(\sqrt[4]{2}\).
For all other \(n>2\), the value of \(\sqrt[n^2]{n}\) will only decrease and will be closing to \(1\).
This will save a lot of time.
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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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My solution is presented attached.
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