题目内容
已知等比数列{an}满足an>0,n=1,2,3…,且a5•a6=8,则log2a2+log2a3+log2a4+log2a5+log2a6+log2a7+log2a8+log2a9=( )
分析:由等比数列的定义和性质可得a2a9=a3a8=a4a7=a5a6=8,要求的式子即log2(a2a9•a3a8•a4a7•a5a6),即log284,再利用对数的运算性质求出结果.
解答:解:由等比数列{an}满足an>0,n=1,2,3…,且a5•a6=8,可得
a2a9=a3a8=a4a7=a5a6=8.
∴log2a2+log2a3+log2a4+log2a5+log2a6+log2a7+log2a8+log2a9 =log2(a2a9•a3a8•a4a7•a5a6)
=log284=log2212=12,
故选D.
a2a9=a3a8=a4a7=a5a6=8.
∴log2a2+log2a3+log2a4+log2a5+log2a6+log2a7+log2a8+log2a9 =log2(a2a9•a3a8•a4a7•a5a6)
=log284=log2212=12,
故选D.
点评:本题主要考查等比数列的定义和性质,对数的运算性质的应用,属于中档题.
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