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=______.
答
由题意可得a5a6+a4a7=2a5a6=18,解得a5a6=9,
∴log3a1+log3a2+…+log3a10=log3(a1a2…a10)
=log3(a5a6)5=log395=log3310=10
故答案为:10