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.
x^y where x>1
[#permalink]
21 Mar 2023, 01:40
21 Mar 2023, 01:40
Expert Reply
1
Bookmarks
00:00
Question Stats:
54% (01:57) correct
45% (01:31) wrong based on 11 sessions
Hide Show timer Statistics
\(x^\sqrt{y}=\sqrt[y]{x}\) , 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
[#permalink]
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
[#permalink]
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
