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If x > 0, x^2=2^64, and x^x=2^y, what is the value of y?
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21 Jul 2020, 02:23
21 Jul 2020, 02:23
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Question Stats:
49% (01:54) correct
49% (01:19) wrong based on 14 sessions
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If \(x > 0\), \(x^2=2^{64}\), and \(x^x=2^y\), what is the value of y?
A. \(2\)
B. \(2^{11}\)
C. \(2^{32}\)
D. \(2^{37}\)
E. \(2^{64}\)
A. \(2\)
B. \(2^{11}\)
C. \(2^{32}\)
D. \(2^{37}\)
E. \(2^{64}\)
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Re: If x > 0, x^2=2^64, and x^x=2^y, what is the value of y?
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21 Jul 2020, 08:34
21 Jul 2020, 08:34
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\(x^2=2^{64}\)
Can be written as
\(x^2=2^{32*2}\) = \({2^{32}}^2\) [ Using \(a^{b*c}\) = \((a^b)^c\) ]
Using "If base of two exponents is same then their power will also be same"
=> \(x = 2^{32}\) [ Using \(a^b\) = \(a^c\) => b = c
\(x^x=2^y\)
=> \(({2^{32}})^{2^{32}} =2^y\)
=> \(2^{{32*2^{32}}} =2^y\) [ Using \((a^b)^c\) = \(a^{b*c}\) ]
Now we know that 32 = \(2^5\)
=> \(2^{2^5*2^{32} =2^y\)
Using if two exponents with same base are multiplied then their power gets added up.
Now \(2^5\) * \(2^{32}\) will become \(2^{5 + 32}\) = \(2^{37}\) [Using \(x^a * x^b = x^{a+b}\) ]
=> \(2^{2^{37}} = 2^y\)
=> y = \(2^{37}\)
So, Answer will be D
Hope it helps!
Can be written as
\(x^2=2^{32*2}\) = \({2^{32}}^2\) [ Using \(a^{b*c}\) = \((a^b)^c\) ]
Using "If base of two exponents is same then their power will also be same"
=> \(x = 2^{32}\) [ Using \(a^b\) = \(a^c\) => b = c
\(x^x=2^y\)
=> \(({2^{32}})^{2^{32}} =2^y\)
=> \(2^{{32*2^{32}}} =2^y\) [ Using \((a^b)^c\) = \(a^{b*c}\) ]
Now we know that 32 = \(2^5\)
=> \(2^{2^5*2^{32} =2^y\)
Using if two exponents with same base are multiplied then their power gets added up.
Now \(2^5\) * \(2^{32}\) will become \(2^{5 + 32}\) = \(2^{37}\) [Using \(x^a * x^b = x^{a+b}\) ]
=> \(2^{2^{37}} = 2^y\)
=> y = \(2^{37}\)
So, Answer will be D
Hope it helps!
Re: If x > 0, x^2=2^64, and x^x=2^y, what is the value of y?
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18 Aug 2023, 04:05
18 Aug 2023, 04:05
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
