求值：(log4(3)+log8(3))(log3(2)+log3(4))-log2(64)

换底公式
原式=(lg3/lg4+lg3/lg8)(lg2/lg3+lg4/lg3)-log2(2^6)
=(lg3/2lg2+lg3/3lg2)(lg2/lg3+2lg2/lg3)-log2(2^6)
=[(1/2+1/3)(lg3/lg2)][(1+2)(lg2/lg3)]-6log2(2)
约分
=(1/2+1/3)(1+2)-6
=5/2-6
=-7/2