log(3)＞log(2)＞0 だから、例えば底をすべて10に揃えると、
log_{3}(√3)=(1/2)*log(3)/log(3)=1/2＜1、log_{3}(√2)=(1/2)*(log(2)/log(3))＜1
log_{1/3}(√(3/2))=(1/2)*log(3/2)/log(1/3)=-(1/2)*{log(3)-log(2)}/log(3)＜0
log_{√3}(√6)=(1/2)*log(6)/{(1/2)*log(3)}={log(2)+log(3)}/log(3)={log(2)/log(3)}+1＞1