已知a>2,求证:log(a-1)a>loga(a+1)
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已知a>2,求证:log(a-1)a>loga(a+1)
证明(法一):∵log(a−1)a−loga(a+1)=
1
loga(a−1)−loga(a+1)
=
1−(loga(a−1))•(loga(a+1))
loga(a−1).
因为a>2,所以,loga(a-1)>0,loga(a+1)>0,
所以,loga(a-1)•loga(a+1)≤[
loga(a−1)+loga(a+1)
2]2
=
[loga(a2−1)]2
4<
[logaa2]2
4=1
所以,log(a-1)a-loga(a+1)>0,命题得证.
证明2:因为a>2,所以,loga(a-1)>0,loga(a+1)>0,
所以,
log(a−1)a
loga(a+1)=
1
loga(a−1)
loga(a−1)=
1
(loga(a−1))•(loga(a+1))
由法1可知:loga(a-1)•loga(a+1)≤[
loga(a−1)+loga(a+1)
2]2
=
[loga(a2−1)]2
4<
[logaa2]2
4=1
∴
1
loga(a−1)•loga(a+1)>1.
故命题得证
1
loga(a−1)−loga(a+1)
=
1−(loga(a−1))•(loga(a+1))
loga(a−1).
因为a>2,所以,loga(a-1)>0,loga(a+1)>0,
所以,loga(a-1)•loga(a+1)≤[
loga(a−1)+loga(a+1)
2]2
=
[loga(a2−1)]2
4<
[logaa2]2
4=1
所以,log(a-1)a-loga(a+1)>0,命题得证.
证明2:因为a>2,所以,loga(a-1)>0,loga(a+1)>0,
所以,
log(a−1)a
loga(a+1)=
1
loga(a−1)
loga(a−1)=
1
(loga(a−1))•(loga(a+1))
由法1可知:loga(a-1)•loga(a+1)≤[
loga(a−1)+loga(a+1)
2]2
=
[loga(a2−1)]2
4<
[logaa2]2
4=1
∴
1
loga(a−1)•loga(a+1)>1.
故命题得证
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