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设等比数列{an}的各项均为正数,且a5a6+a4a7=18,则log3a1+log3a2+…+log3a10=______.
人气:313 ℃ 时间:2020-05-11 13:08:23
解答
由题意可得a5a6+a4a7=2a5a6=18,解得a5a6=9,
∴log3a1+log3a2+…+log3a10=log3(a1a2…a10
=log3(a5a65=log395=log3310=10
故答案为:10
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