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The operator ! is defined such that a!b = a^b
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16 Sep 2017, 15:40
16 Sep 2017, 15:40
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Question Stats:
86% (01:13) correct
13% (02:13) wrong based on 87 sessions
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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.
Re: The operator ! is defined such that a!b = a^b
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28 Sep 2017, 08:13
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!
Thus, answer C!
Re: The operator ! is defined such that a!b = a^b
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02 Jan 2022, 07:30
02 Jan 2022, 07:30
Expert Reply
