等比数列，若a5*a6=81，求log3a1+log3a2+....+log3a10的值

来源：百度知道 编辑：UC知道 时间：2024/06/11 02:53:31
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解:
由于{an}为等比数列
则:a5a6=a4a7=a3a8=a2a9=a1a10

又 a5a6=81

则:
log3(a1)+log3(a2)+...+log3(a9)+log3(a10)
=log3[a1*a2*a3*...*a10]
=log3[(a1a10)*(a2a9)*...*(a5a6)]
=log3[81*81*...*81]
=log3[81^5]
=log3[3^20]
=20log3[3]
=20

log3a1+log3a2+....+log3a10
=(log3a1+log3a10)+(log3a2+log3a9)+...+(log3a5+log3a6)
=log3a1a10+log3a2a9+...+log3a5a6
=5log3a5a6
=5log3(81)
=5*4
=20

a5*a6=81
a1*q^5*a1*q^6=81
a1^2*q^9=81
log3a1+log3a2+....+log3a10
=log3(a1*a2*...a10)
=log3(a1^10*q^ 45)
=log3(a1^2*q^9)^5
=5log3 81
=20