题目内容
设0<x<1,a>0且a≠| 1 | 3 |
分析:分类讨论,当3a>1和当0<3a<1两种情况,去掉绝对值,化简这2个式子,比较它们与0的大小关系.
解答:解:∵0<x<1,分2种情况讨论,
①当3a>1,即a>
时,
|log3a(1-x)3|-|log3a(1+x)3|
=|3log3a(1-x)|-|3log3a(1+x)|
=3[-log3a(1-x)-log3a(1+x)]
=-3log3a(1-x2).
∵0<1-x2<1,∴-3log3a(1-x2)>0.
②当0<3a<1,即0<a<
时,
|log3a(1-x)3|-|log3a(1+x)3|
=3[log3a(1-x)+log3a(1+x)]
=3log3a(1-x2)>0.
综上所述,|log3a(1-x)3|>|log3a(1+x)
①当3a>1,即a>
| 1 |
| 3 |
|log3a(1-x)3|-|log3a(1+x)3|
=|3log3a(1-x)|-|3log3a(1+x)|
=3[-log3a(1-x)-log3a(1+x)]
=-3log3a(1-x2).
∵0<1-x2<1,∴-3log3a(1-x2)>0.
②当0<3a<1,即0<a<
| 1 |
| 3 |
|log3a(1-x)3|-|log3a(1+x)3|
=3[log3a(1-x)+log3a(1+x)]
=3log3a(1-x2)>0.
综上所述,|log3a(1-x)3|>|log3a(1+x)
点评:本题考查对数的运算性质,体现分类讨论的数学思想.
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