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

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The operator ! is defined such that a!b = a^b [#permalink]  16 Sep 2017, 15:40
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Question Stats:

90% (00:51) correct 10% (05:52) wrong based on 10 sessions

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.
[Reveal] Spoiler: OA

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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.

Re: The operator ! is defined such that a!b = a^b   [#permalink] 28 Sep 2017, 08:13
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# The operator ! is defined such that a!b = a^b

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