设等比数列{an}的各项均为正数,且a5a6+a4a7=18,则log3a1+log3a2+…+log3a10=______.

问题描述:

设等比数列{an}的各项均为正数,且a5a6+a4a7=18,则log3a1+log3a2+…+log3a10=______.

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
答案解析:由题意可得a4a7=a5a6,解之可得a5a6,由对数的运算可得log3a1+log3a2+…+log3a10=log3(a1a2…a10)=log3(a5a65,代入计算可得.
考试点:等比数列的性质.
知识点:本题考查等比数列的性质和通项公式,涉及对数的运算,属中档题.