在等比数列{an}中,a4a5=32,log2a1+loga2+…+log2a8=______.
人气:447 ℃ 时间:2019-10-10 00:46:59
解答
正项等比数列{an}中,
∵log2a1+log2a2+…+log2a8 =log2[a1a8•a2a7•a3a6•a4a5]=log2(a4a5)4
=log2324=20,
故答案为:20
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