GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
The operator ! is defined such that a!b = a^b
[#permalink]
16 Sep 2017, 15:40
16 Sep 2017, 15:40
Expert Reply
00:00
Question Stats:
85% (01:12) correct
14% (02:15) wrong based on 92 sessions
Hide Show timer Statistics
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
[#permalink]
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
[#permalink]
02 Jan 2022, 07:30
02 Jan 2022, 07:30
Expert Reply
