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16 Sep 2014, 01:24
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If \(x \gt 0\), \(x^2=2^{64}\), 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. \(2^{64}\)
16 Sep 2014, 01:24
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Official Solution:
If \(x \gt 0\), \(x^2=2^{64}\), and \(x^x=2^y\), what is the value of \(y\)?
A. 2
B. \(2^{11}\)
C. \(2^{32}\)
D. \(2^{37}\)
E. \(2^{64}\)
Given that \(x^2 = 2^{64}\), we can deduce that \(x = \sqrt{2^{64} } = 2^{\frac{64}{2} } = 2^{32}\). (Note: The solution \(x = -\sqrt{2^{64} }\) is not valid because we are given that \(x > 0\)).
For the second step: \(x^x = (2^{32})^{(2^{32})} = 2^{32*2^{32} } = 2^{2^{5}*2^{32} } = 2^{2^{37} } = 2^y\). So, we have that \(2^{2^{37} } = 2^y\), which means that \(y = 2^{37}\).
Alternatively: \(x^x = (2^{32})^x = 2^{32x} = 2^y\). So, we have that \(2^{32x} = 2^y\), which implies \(y = 32x\). Given that \(x = 2^{32}\), then \(y = 32 * 2^{32} = 2^5 * 2^{32} = 2^{37}\).
Answer: D
If \(x \gt 0\), \(x^2=2^{64}\), and \(x^x=2^y\), what is the value of \(y\)?
A. 2
B. \(2^{11}\)
C. \(2^{32}\)
D. \(2^{37}\)
E. \(2^{64}\)
Given that \(x^2 = 2^{64}\), we can deduce that \(x = \sqrt{2^{64} } = 2^{\frac{64}{2} } = 2^{32}\). (Note: The solution \(x = -\sqrt{2^{64} }\) is not valid because we are given that \(x > 0\)).
For the second step: \(x^x = (2^{32})^{(2^{32})} = 2^{32*2^{32} } = 2^{2^{5}*2^{32} } = 2^{2^{37} } = 2^y\). So, we have that \(2^{2^{37} } = 2^y\), which means that \(y = 2^{37}\).
Alternatively: \(x^x = (2^{32})^x = 2^{32x} = 2^y\). So, we have that \(2^{32x} = 2^y\), which implies \(y = 32x\). Given that \(x = 2^{32}\), then \(y = 32 * 2^{32} = 2^5 * 2^{32} = 2^{37}\).
Answer: D
General Discussion
28 Aug 2023, 14:47
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
