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Which is greater (1/x+5)^5 or (1/x+5)^3
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Updated on: 16 Dec 2018, 09:23
Updated on: 16 Dec 2018, 09:23
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Question Stats:
66% (00:36) correct
33% (00:42) wrong based on 18 sessions
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\(X < -1\)
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
\((\frac{1}{x+5})^5\) |
\((\frac{1}{x+5})^3\) |
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Re: Fraction
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16 Dec 2018, 07:49
16 Dec 2018, 07:49
1
Expert Reply
Sawant91 wrote:
X < -1
(1/x+5)^5
(1/x+5)^3
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A
(1/x+5)^5
Quantity B
(1/x+5)^3
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
The moment you see that x<-1 and the denominator is x+5, you should realize that x+5 can be both positive, when x>-5 and negative when x<-5... so answer would differ in each case..
let x = -2..
A=\((\frac{1}{x+5})^5=(\frac{1}{-2+5})^5=(\frac{1}{3})^5=\frac{1}{243}\)...
B=\((\frac{1}{x+5})^3=(\frac{1}{-2+5})^3=(\frac{1}{3})^3=\frac{1}{27}\)
so B>A
let x = -6..
A=\((\frac{1}{x+5})^5=(\frac{1}{-6+5})^5=(\frac{1}{-1})^5=-1\)...
B=\((\frac{1}{x+5})^3=(\frac{1}{-6+5})^3=(\frac{1}{-1})^3=-1\)
so B=A
Thus relation cannot be determined....
D
Re: Which is greater (1/x+5)^5 or (1/x+5)^3
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16 Dec 2018, 09:24
16 Dec 2018, 09:24
Expert Reply
Please follow the rules for posting on the board https://gre.myprepclub.com/forum/rules-for ... -1083.html
and the way (very simple) to format a question https://gre.myprepclub.com/forum/qq-how-to ... -2357.html
Thank you so much
and the way (very simple) to format a question https://gre.myprepclub.com/forum/qq-how-to ... -2357.html
Thank you so much
