求值(log43+log83)(log32+log92) 注:log后面那为数是底数
问题描述:
求值(log43+log83)(log32+log92) 注:log后面那为数是底数
=5/6*log3/log2*3/2*log2/log3
=5/4 怎么来的
答
(log43+log83)(log32+log92)=(lg3/2lg2+;lg3/3lg2)(lg2/lg3+lg2/2lg3) (换底公式)
=lg3/lg2(1/2+1/3)(lg2/lg3(1+1/2)
=5/6x3/2
=5/4