解:(1)∵y=log
2x在(0,+∞)上為增函數(shù),
又∵3<5,∴l(xiāng)og
23<log
25.
又∵y=0.6
x在R上為減函數(shù),
∴
![](http://thumb.zyjl.cn/pic5/latex/92561.png)
.
(2)∵log
aa=1,a>0且a≠1
①當(dāng)a>1時(shí),a
2>1,∴
![](http://thumb.zyjl.cn/pic5/latex/92562.png)
;
②當(dāng)0<a<1時(shí),a
2<1,∴
![](http://thumb.zyjl.cn/pic5/latex/92563.png)
.
分析:(1)利用指數(shù)函數(shù)和對(duì)數(shù)函數(shù)的單調(diào)性即可比較出其大��;
(2)利用對(duì)數(shù)函數(shù)和冪函數(shù)的單調(diào)性即可比較出大小.
點(diǎn)評(píng):熟練掌握指數(shù)函數(shù)和對(duì)數(shù)函數(shù)即冪函數(shù)的單調(diào)性是解題的關(guān)鍵.