Question

9、將2^0.3，log_0.32，log_0.33三個(gè)數(shù)按從小到大的順序排列為

log_0.33＜log_0.32＜2^0.3

．

Answer 1

分析：根據(jù)指數(shù)和對(duì)數(shù)的性質(zhì)可得2^0.3大于0，log_0.3²和log_0.3³都小于0，然后利用對(duì)數(shù)底數(shù)0.3＜1，根據(jù)對(duì)數(shù)函數(shù)為減函數(shù)得到大小即可．

解答：解：由指數(shù)和對(duì)數(shù)函數(shù)的性質(zhì)得：2^0.3＞0，log_0.3²＜0，log_0.3³＜0；
而y=log_0.3^x為底數(shù)是0.3＜1的對(duì)數(shù)函數(shù)且是減函數(shù)，
由3＞2得到，log_0.3²＞log_0.3³，
所以三個(gè)數(shù)的大小順序?yàn)閘og_0.33＜log_0.32＜2^0.3
故答案為log_0.33＜log_0.32＜2^0.3

點(diǎn)評(píng)：考查學(xué)生靈活運(yùn)用指數(shù)和對(duì)數(shù)函數(shù)的性質(zhì)及利用對(duì)數(shù)函數(shù)的增減性比較大小，學(xué)生做題時(shí)應(yīng)利用函數(shù)思想進(jìn)行比較大�。�