分析:本題所給的不等式是一個(gè)對(duì)數(shù)不等式,我們要先利用換底公式將不等式的二項(xiàng)均化為同底,再根據(jù)對(duì)數(shù)函數(shù)的單調(diào)性,即可得到答案.
解答:解:不等式果 log
a3>log
b3>0,可化為:
>>0,
故0<log
3a<log
3b
又∵函數(shù)y=log
3x的底數(shù)3>1,
故函數(shù)y=log
3x為增函數(shù)
∴1<a<b,
故答案為:B
點(diǎn)評(píng):本題考查的知識(shí)點(diǎn)是對(duì)數(shù)函數(shù)的單調(diào)性與特殊點(diǎn),其中根據(jù)對(duì)數(shù)函數(shù)的性質(zhì)將對(duì)數(shù)不等式轉(zhuǎn)化為一個(gè)整式不等式是解答本題的關(guān)鍵.要注意對(duì)數(shù)函數(shù)的單調(diào)性,即當(dāng)?shù)讛?shù)大于1時(shí)單調(diào)遞增,當(dāng)?shù)讛?shù)大于0小于1時(shí)單調(diào)遞減.