If x = log_3 5 and y = log_{17} 25,then which | vedclass

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(C) Given $x = \log_3 5$ and $y = \log_{17} 25 = 2 \log_{17} 5$.Taking reciprocals:$\frac{1}{x} = \log_5 3 = \log_5 (9^{1/2}) = \frac{1}{2} \log_5 9$.

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$x > y$

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(C) Given $x = \log_3 5$ and $y = \log_{17} 25 = 2 \log_{17} 5$.Taking reciprocals:$\frac{1}{x} = \log_5 3 = \log_5 (9^{1/2}) = \frac{1}{2} \log_5 9$.$\frac{1}{y} = \frac{1}{2} \log_5 17$.Since $17 > 9$,it follows that $\log_5 17 > \log_5 9$.Therefore,$\frac{1}{2} \log_5 17 > \frac{1}{2} \log_5 9$,which implies $\frac{1}{y} > \frac{1}{x}$.Since $x$ and $y$ are both positive,$\frac{1}{y} > \frac{1}{x}$ implies $x > y$.

If $x = \log_3 5$ and $y = \log_{17} 25$,then which of the following is correct?

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