Question 1040181
To prove the statement, first note that {{{log(a,b)*log(b,a) = 1}}}, since 
{{{log(a,b) = (log(b,b))/(log(b,a)) = 1/(log(b,a)) }}}, by direct use of the change-of-base formula.

Thus...

{{{log(a,x)*log(b,y) = ((log(b,x))/(log(b,a))  )*((log(a,y))/(log(a,b))) = ((log(b,x))*(log(a,y)))/( log(a,b)*log(b,a)) = log(b,x)*log(a,y)}}}