For , let then the value of k can NEVER be equal to

1. . Let . Since , . We know for all . Therefore, . Hence, can NEVER be .

For x\ge y \gt 1, let \log_{x}(\frac{x}{y})+\log_{y}(\frac{y}{x})=k, then the value of k can <strong>NEVER</strong> be equal to

A. -1
B. -\frac{1}{2}
C. 0
D. 1 ✓

Correct Answer: D. 1

Explanation

k = (\log_x x - \log_x y) + (\log_y y - \log_y x) = 2 - (\log_x y + \log_y x). Let t = \log_x y. Since x \geq y \gt 1, 0 \lt t \leq 1. We know t + \frac{1}{t} \geq 2 for all t \gt 0. Therefore, k = 2 - (t + \frac{1}{t}) \leq 0. Hence, k can <strong>NEVER</strong> be 1.

For x\ge y \gt 1, let \log_{x}(\frac{x}{y})+\log_{y}(\frac{y}{x})=k, then the value of k can <strong>NEVER</strong> be equal to

Explanation

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