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Let x=(1/3)^-1/2, y=(1/2)^-1/3, and z=(1/4)^-1/4.

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Let x=(1/3)^-1/2, y=(1/2)^-1/3, and z=(1/4)^-1/4. [#permalink]

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New post 27 Mar 2018, 02:19
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[GMAT math practice question]

Let \(x=(\frac{1}{3})^{\frac{-1}{2}}\), \(y=(\frac{1}{2})^{-\frac{1}{3}}\), and \(z=(\frac{1}{4})^{\frac{-1}{4}}\). Which of the following is true?

\(A. x<y<z\)
\(B. z<y<x\)
\(C. x<z<y\)
\(D. y<z<x\)
\(E. y<x<z\)
[Reveal] Spoiler: OA

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Re: Let x=(1/3)^-1/2, y=(1/2)^-1/3, and z=(1/4)^-1/4. [#permalink]

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New post 27 Mar 2018, 07:49
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MathRevolution wrote:
[GMAT math practice question]

Let \(x=(\frac{1}{3})^\frac{-1}{2}\), \(y=(\frac{1}{2})^-\frac{1}{3}\), and \(z=(\frac{1}{4})^\frac{-1}{4}\). Which of the following is true?

\(A. x<y<z\)
\(B. z<y<x\)
\(C. x<z<y\)
\(D. y<z<x\)
\(E. y<x<z\)




\(x=(\frac{1}{3})^\frac{-1}{2}\)
= \(1/\frac{1}{3}^\frac{1}{2}\)
= \(1/\frac{1}{\sqrt{3}}\)
= \(\sqrt{3}\) = 1.7 approx

Similarly , \(y=(\frac{1}{2})^-\frac{1}{3}\) = \(\sqrt[3]{2}\) = 1.2 approx

And, and \(z=(\frac{1}{4})^\frac{-1}{4}\) = \(\sqrt[4]{4}\) = \(\sqrt[2]{2}\) =1.4 approx

Hence , y<z<x ans D
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Re: Let x=(1/3)^-1/2, y=(1/2)^-1/3, and z=(1/4)^-1/4. [#permalink]

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New post 28 Mar 2018, 12:00
As nobody could really calculate the squares you just have to simplify them:

x = (1/3)^(-1/2) = 3^(1/2)
y = (1/2)^(-1/3) = 2^(1/3)
z = (1/4)^(-1/4) = 4^(1/4) = 2^(1/2)

I think now it's easy to sort them by size.
Expert Post
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Posts: 5255
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GPA: 3.82
Re: Let x=(1/3)^-1/2, y=(1/2)^-1/3, and z=(1/4)^-1/4. [#permalink]

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New post 29 Mar 2018, 01:12
=>
We raise each of \(x, y\), and \(z\) to the exponent \(12\) as this will yield integers that are easily compared.
\(x^{12} = ((\frac{1}{3})^{\frac{-1}{2}})^{12} = ((3)^{\frac{1}{2}})^{12} = 3^6 = 729\)
\(y^{12} = ((\frac{1}{2})^{\frac{-1}{3}})^{12} = ((2)^{\frac{1}{3}})^{12} = 2^4 = 16\)
\(z^{12} = ((\frac{1}{4})^{\frac{-1}{4}})^{12} = ((4)^{\frac{1}{4}})^{12} = 4^3 = 64\)

We have \(y^{12} < z^{12} < z^{12}\). Since \(x, y\), and \(z\) are all positive, this tells use that \(y < z < x.\)

Therefore, the answer is D.

Answer: D
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Re: Let x=(1/3)^-1/2, y=(1/2)^-1/3, and z=(1/4)^-1/4.   [#permalink] 29 Mar 2018, 01:12
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Let x=(1/3)^-1/2, y=(1/2)^-1/3, and z=(1/4)^-1/4.

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