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The operator ! is defined such that a!b = a^b

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Carcass

Verbal Expert

Joined: 18 Apr 2015

Posts: 21063

The operator ! is defined such that a!b = a^b [#permalink]

16 Sep 2017, 15:40

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This question is part of GREPrepClub - The Questions Vault Project

The operator ! is defined such that \(a!b = a^b * b^{-a}\)

Quantity A	Quantity B
\(\frac{x!4}{4!x}\)	\(\frac{x^8}{16^x}\)

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

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IlCreatore

Director

Joined: 03 Sep 2017

Posts: 518

Re: The operator ! is defined such that a!b = a^b [#permalink]

28 Sep 2017, 08:13

Column A is \(\frac{x^44^{-x}}{4^xx^{-4}}\). Then using the properties of exponentials, \(\frac{x^4}{x^{-4}} = x^8\) and \(\frac{4^{-x}}{4^x} = \frac{1}{4^{2x}} = \frac{1}{16^x}\). Thus, column A can be rewritten as \(\frac{x^8}{16^x}\) that is exactly the expression of column B.

Thus, answer C!