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16 Sep 2014, 01:26
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If \(x \gt 0\), what is the value of \(y^x\)?
(1) \(\frac{4^{(x+y)^2}}{4^{(x-y)^2}}=128^{xy}\)
(2) \(x \ne 1\) and \(x^y=1\)
(1) \(\frac{4^{(x+y)^2}}{4^{(x-y)^2}}=128^{xy}\)
(2) \(x \ne 1\) and \(x^y=1\)
16 Sep 2014, 01:26
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Official Solution:
If \(x \gt 0\), what is the value of \(y^x\)?
(1) \(\frac{4^{(x+y)^2}}{4^{(x-y)^2}}=128^{xy}\).
Simplifying this equation, we get \(4^{(x+y)^2-(x-y)^2}=128^{xy}\).
Expanding and simplifying the exponent of 4 gives \(4^{4xy}=128^{xy}\).
Expressing each side in terms of powers of 2 gives \(2^{8xy}=2^{7xy}\);
Equating the powers, we find \(8xy=7xy\), which implies \(xy=0\).
Since given that \(x > 0\), we can deduce that \(y=0\), and therefore \(y^x=0^x=0\).
Sufficient.
(2) \(x \ne 1\) and \(x^y=1\).
Since \(x > 0\) and \(x \ne 1\), the only possibility for \(x^y=1\) to hold true is when \(y=0\). Thus, \(y^x=0^x=0\). Sufficient.
Answer: D
If \(x \gt 0\), what is the value of \(y^x\)?
(1) \(\frac{4^{(x+y)^2}}{4^{(x-y)^2}}=128^{xy}\).
Simplifying this equation, we get \(4^{(x+y)^2-(x-y)^2}=128^{xy}\).
Expanding and simplifying the exponent of 4 gives \(4^{4xy}=128^{xy}\).
Expressing each side in terms of powers of 2 gives \(2^{8xy}=2^{7xy}\);
Equating the powers, we find \(8xy=7xy\), which implies \(xy=0\).
Since given that \(x > 0\), we can deduce that \(y=0\), and therefore \(y^x=0^x=0\).
Sufficient.
(2) \(x \ne 1\) and \(x^y=1\).
Since \(x > 0\) and \(x \ne 1\), the only possibility for \(x^y=1\) to hold true is when \(y=0\). Thus, \(y^x=0^x=0\). Sufficient.
Answer: D
General Discussion
02 Feb 2016, 09:22
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Hi
how is xy=0?; is it because only way 7xy=8xy is when xy=0?
how is xy=0?; is it because only way 7xy=8xy is when xy=0?
