(log4上面3+log8上面3)乘以(log3上面2+log9上面2)—log上面4次根号32=?
1个回答
展开全部
(log4(3)+log8(3))(log3(2)+log9(2))—log4(32)^1/4
=(1/2log2(3)+1/3log2(3))(log3(2)+1/2log3(2))-1/4*2log2(32)
=5/6log2(3)*3/2log3(2)-1/2log2(32)
=5/6*3/2*6log2(3)*2log3(2)-1/2[log2(8)+log2(4)]
=5/4-1/2[log2(2)^3+log2(2)^2]
=5/4-1/2(3+2)
= -5/4
=(1/2log2(3)+1/3log2(3))(log3(2)+1/2log3(2))-1/4*2log2(32)
=5/6log2(3)*3/2log3(2)-1/2log2(32)
=5/6*3/2*6log2(3)*2log3(2)-1/2[log2(8)+log2(4)]
=5/4-1/2[log2(2)^3+log2(2)^2]
=5/4-1/2(3+2)
= -5/4
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询