已知{an}为等比数列,an>0,且a1a2009=22010,则log2a1+log2a3+…+log2a2009=A.10042B.10052C.10062D.1004×1005
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B
解析分析:由等比数列的性质:a1a2009=a3a2007=…=a1003a1007=a10052=22010,再由对数的运算法则求解即可.
解答:由等比数列的性质:a1a2009=a3a2007=…=a1003a1007=a10052=22010,而log2a1+log2a3+…+log2a2009=log2a1?a3…a2009=10052故选B.
点评:本题考查等比数列的性质的应用和对数的运算法则,属基础知识、基本运算的考查.