\(x = \log_3 5 > 1\), since \(5 > 3\). In fact \(x \approx 1.465\).

\(y = \log_{17}25 = \frac{\ln 25}{\ln 17} \approx \frac{3.219}{2.833} \approx 1.136\).

So \(x > y\).

(Without a calculator: \(x = \log_3 5 > \log_3\sqrt{27} = 1.5\) is false. But \(3^{1.4} \approx 4.66 < 5\) gives \(x > 1.4\), and \(17^{1.2} \approx 30 > 25\) gives \(y < 1.2\).)

Answer: (A) x > y