已知01,则下列不等式中成立的是（ ）A.log(b)(1/b)＜log(a)(b)＜log(a)(1/b)B.log(a)(b)＜log(b)(1/b)＜log(a)(1/b)C.log(a)(b)＜log(a)(1/b)＜log(b)(1/b)D.log(b)(1/b)＜log(a)(1/b)＜log(a)(b)

已知01,则下列不等式中成立的是（ ）A.log(b)(1/b)＜log(a)(b)＜log(a)(1/b)B.log(a)(b)＜log(b)(1/b)＜log(a)(1/b)C.log(a)(b)＜log(a)(1/b)＜log(b)(1/b)D.log(b)(1/b)＜log(a)(1/b)＜log(a)(b)
已知01,则下列不等式中成立的是（ ）
A.log(b)(1/b)＜log(a)(b)＜log(a)(1/b)
B.log(a)(b)＜log(b)(1/b)＜log(a)(1/b)
C.log(a)(b)＜log(a)(1/b)＜log(b)(1/b)
D.log(b)(1/b)＜log(a)(1/b)＜log(a)(b)

已知01,则下列不等式中成立的是（ ）A.log(b)(1/b)＜log(a)(b)＜log(a)(1/b)B.log(a)(b)＜log(b)(1/b)＜log(a)(1/b)C.log(a)(b)＜log(a)(1/b)＜log(b)(1/b)D.log(b)(1/b)＜log(a)(1/b)＜log(a)(b)
选B.
01 即b>1/a 由log(a)(x)单调递减,所以
log(a)(b)0
所以log(a)(b)＜log(b)(1/b)＜log(a)(1/b)
所以B成立.

令a=1/2,b=4,带入验证：
log(b)(1/b)=-1
log(a)(b)=-2
log(a)(1/b)=2
所以B log(a)(b)＜log(b)(1/b)＜log(a)(1/b)成立