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x^y where x>1
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21 Mar 2023, 01:40
21 Mar 2023, 01:40
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Question Stats:
55% (02:15) correct
44% (01:21) wrong based on 9 sessions
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\(x^\sqrt{y}=\sqrt[y]{x}\) , where x >1
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.
Quantity A |
Quantity B |
y |
1 |
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: x^y where x>1
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21 Mar 2023, 05:16
21 Mar 2023, 05:16
Carcass wrote:
\(x^y=\sqrt[y]{x}\) , where x >1
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.
Quantity A |
Quantity B |
y |
1 |
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.
Can someone please explain?
Re: x^y where x>1
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21 Mar 2023, 05:43
21 Mar 2023, 05:43
1
Expert Reply
Given that \(x^{\sqrt{y}}=\sqrt[y]{x }\), where \(x>1\)
\(x^{\sqrt{y}}=\sqrt[y]{x }\)
=> \(x^{\sqrt{y}}= x^{\frac{1}{y}}\)
Since the base of the two exponents is same (x) => power will also be the same
=> \(\sqrt{y}\) = \(\frac{1}{y}\)
=> \(y*\sqrt{y}\) = 1
=> \(y ^ {1 + \frac{1}{2}}\) = 1
=> \(y ^ {\frac{3}{2}}\) = 1
=> y = \(1^{\frac{2}{3}}\) = 1
Clearly, Quantity A = Quantity B = 1
So, Answer will be C
The solution is also located here https://gre.myprepclub.com/forum/gre-mi ... ml#p105811
\(x^{\sqrt{y}}=\sqrt[y]{x }\)
=> \(x^{\sqrt{y}}= x^{\frac{1}{y}}\)
Since the base of the two exponents is same (x) => power will also be the same
=> \(\sqrt{y}\) = \(\frac{1}{y}\)
=> \(y*\sqrt{y}\) = 1
=> \(y ^ {1 + \frac{1}{2}}\) = 1
=> \(y ^ {\frac{3}{2}}\) = 1
=> y = \(1^{\frac{2}{3}}\) = 1
Clearly, Quantity A = Quantity B = 1
So, Answer will be C
The solution is also located here https://gre.myprepclub.com/forum/gre-mi ... ml#p105811
