Question

已知a>2,试判断loga(a-1)*loga(a+1)与1的大小关系 要有过程

Answer 1

因为 ab<=(a+b)^2/4
所以
loga(a-1)*loga(a+1)
<=[loga(a-1)+loga(a+1)]^2/4
=[loga(a^2-1)]^2/4
<[loga(a^2)]^2/4
=[2log(a,a)]^2/4
=2^2/4=1
从而
loga(a-1)*loga(a+1)<1.

Answer 2

根据题意，loga (a-1)>0，loga (a+1)>0，且loga x是一个增函数。则有：
[loga (a-1)]*[loga (a+1)]
<{[loga (a-1)+loga (a+1)]/2}^2
=[loga (a^2-1)]^2/4
<(loga a^2)^2/4
=2^2/4
=1.
不等式等号成立的条件是loga (a-1)=loga (a+1)【不可能】，所以只取小于号。
综上可知，[loga (a-1)]*[loga (a+1)]<1.【注：此题用到了基本不等式ab<=[(a+b)/2]^2，a、b为正数】

Answer 3

因√[loga(a-1)*loga(a+1)]
<(1/2)[loga (a-1)+loga (a+1)]
=(1/2)loga (a²-1)
<(1/2)loga a²
=1
所以loga(a-1)*loga(a+1)<1