问一道高一数学题,谢谢!!!!

a5a6=81
则a1a11=a2a9=a3a8=a4a7=a5a6=81=3^4

log3a1+log3a2+...+log3a10
=log3(a1*a2*……*a10)
=log3(3^4)^5
=log3 3^20
=20

30log3

的确，如果3为底的话是20。

在各项都为正数的等比数列{an}中
a5<a6
a5*a6=81

log3a1+log3a2+...+log3a10
=log3(a1*a2*a3...*a10)
=log3(a1*a11*a2*a9...*a5*a6)
=log3(81*81*81*81*81)
=20

a5*a6=a1*a10=a2*a9=...=a4*a7=81
原=（log3a1+log3a10）+（log3a2+log3a9）+....+(log3a5+log3a6)=5*log3(81)=20

log3a1+log3a2+log3a3+…+log3a10=log3a1*a2*a3…a10
又a5*a6=81,则a1*a10=81
所以原式=20

an = a1 * q^(n-1)
a5*a6= a1*q^4*a1*q^5 = a1^2*q^9 = 81
log3 a1 + log3 a2 + ...+ log3 a10
= log3 (a1*a2*...*a10)
= log3 (a1*a1^q*a1^2*...a1*q^9)
= log3 ( a1^10 * q^45 )
= log3 ( 81^5)
= log3 ( 3^4^5 )
= log3 3^20
= 20

答案为20