If Log27 in base5=x and Log5 in base3=y., find xy

x = log5 (27)

5^x =27 = 3^3
log3 (3^3) = log3 (5^x)
3 log 3 (3) = x log3 (5)
but log 3 (3) = 1
so
3 = x log3 (5)
log3 (5) = 3/x

and

y = log3 (5)
so
y = 3/x
x y = 3