已知a>2,求证log以a为底(a-1)的对数*以a为底(a+1)的对数<1

2025-05-25 17:35:10
推荐回答(4个)
回答1:

LOGa(a-1)*LOGa(a+1)
<=[LOGa(a-1)+LOGa(a+1)]^2/4 (基本不等式)
=[LOGa(a^2-1)]^2/4
<[LOGa(a^2)]^2/4
=1

回答2:

高中不要求掌握对数的乘法

回答3:

就像楼上的那样做

回答4:

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
=1