题目内容
设等比数列an中,每项均是正数,且a5a6=81,则 log3a1+log3a2+…+log3a10=______.
∵等比数列{an}中,每项均是正数,且a5a6=81,
∴log3a1+log3a2+…+log3a10=log3(a1?a2?a3…a10)
=log3(a5a6)5
=log3320
=20.
故答案:20.
∴log3a1+log3a2+…+log3a10=log3(a1?a2?a3…a10)
=log3(a5a6)5
=log3320
=20.
故答案:20.

练习册系列答案
相关题目