log2(6)*log3(6)-[log2(3)+log3(2)]的值
网友回答
loga(b)=1/logb(a)
原式=log2(2*3)*log3(2*3)-[log2(3)+log3(2)]
=[log2(2)+log2(3)]*[(log3(2)+log3(3)]-[log2(3)+log3(2)]
=[1+log2(3)]*[(log3(2)+1]-[log2(3)+log3(2)]
=log3(2)+1+log2(3)*log3(2)+log2(3)-log2(3)-log3(2)
=log2(3)log3(2)+1
=2