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If \(x\) non-zero integer, how many values of \(x\) are there such that \((\frac{x}{2})^x = 1\)?
A. None
B. 1
C. 2
D. 3
E. Infinitely many
A. None
B. 1
C. 2
D. 3
E. Infinitely many
05 Jun 2024, 10:22
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Official Solution:
If \(x\) non-zero integer, how many values of \(x\) are there such that \((\frac{x}{2})^x = 1\)?
A. None
B. 1
C. 2
D. 3
E. Infinitely many
Since \(x \neq 0\), for \((\frac{x}{2})^x = 1\) to hold true, the base (\(\frac{x}{2}\)) must be 1, and the exponent (\(x\)) can be any number (\(1^{anything} = 1\)) or the base (\(\frac{x}{2}\)) must be -1 and the exponent (\(x\)) must be even (\((-1)^{even} = 1\)).
If \(\frac{x}{2} = 1\), then \(x = 2\).
If \(\frac{x}{2} = -1\), then \(x = -2\), which is even.
Therefore, \(x\) could be 2 or -2.
Answer: C
If \(x\) non-zero integer, how many values of \(x\) are there such that \((\frac{x}{2})^x = 1\)?
A. None
B. 1
C. 2
D. 3
E. Infinitely many
Since \(x \neq 0\), for \((\frac{x}{2})^x = 1\) to hold true, the base (\(\frac{x}{2}\)) must be 1, and the exponent (\(x\)) can be any number (\(1^{anything} = 1\)) or the base (\(\frac{x}{2}\)) must be -1 and the exponent (\(x\)) must be even (\((-1)^{even} = 1\)).
If \(\frac{x}{2} = 1\), then \(x = 2\).
If \(\frac{x}{2} = -1\), then \(x = -2\), which is even.
Therefore, \(x\) could be 2 or -2.
Answer: C
