题目内容
已知a>2,求证:log(a-1)a>loga(a+1).
证明:log(a-1)a-loga(a+1)=
-loga(a+1)=
,
∵a>1,∴loga(a-1)>0,loga(a+1)>0.
∴loga(a-1)·loga(a+1)<[
]2
=
=1.
∴log(a-1)a-loga(a+1)>0,
即log(a-1)a>loga(a+1).
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题目内容
已知a>2,求证:log(a-1)a>loga(a+1).
证明:log(a-1)a-loga(a+1)=
-loga(a+1)=
,
∵a>1,∴loga(a-1)>0,loga(a+1)>0.
∴loga(a-1)·loga(a+1)<[
]2
=
=1.
∴log(a-1)a-loga(a+1)>0,
即log(a-1)a>loga(a+1).