解log3(2)-log4(3)
=lg(2)/lg(3)-lg(3)/lg(4)
=lg(2)lg(4)/lg(3)lg(4)-lg(3)lg(3)/lg(4)lg(3)
=[lg(2)lg(4)-lg(3)lg(3)]/lg(4)lg(3)
而lg(2)lg(4)/lg(3)lg(3)
=log3(2)log3(4)
≤[(log3(2)+log3(4))/2]^2
=[log3(8)/2]^2
＜[log3(9)/2]^2
=1
即lg(2)lg(4)/lg(3)lg(3)＜1
即lg(2)lg(4)＜lg(3)lg(3)
即lg(2)lg(4)-lg(3)lg(3)＜0
即[lg(2)lg(4)-lg(3)lg(3)]/lg(4)lg(3)＜0
即log3(2)＜log4(3)