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