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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] New post 16 Sep 2017, 15:40
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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] New post 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!
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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