Question 1006823
By change-of-base formula,


*[tex \large \log_9 5 = \frac{\log 5}{\log 9}]


So (x,y) = (5,9) works.


But since *[tex \large \log (a^k) = k \log a], we have that *[tex \large (x,y) = (5^k, 9^k)] for any k also works.