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若等比数列An满足log3A1 log3A2......... log3A10=...

log3a1+log3a2+…+log3a10=log3(a1a10)+log3(a2a9)+…log3(a5a6)=5log3(a5a6)=10故答案为:10

a5a6=a4a7=a3a8=a2a9=a1a10=9(正数的等比数列有这性质)好好想想就知道 后面为log3a1a2a3a4a5a6a7a8a9a10=~~~不用我算了吧

解答: 这个是对数的运算法则 同底对数相加,底数不变,真数相乘, 即 loga(M)+loga(N)=loga(MN) 可以推广到有限多个。

a5=a1q^4 q^4=2^4 q=2 sn=36(2^n-1) log3a1+log3a2+…+log3a10=20 log3a1a2a3……a10=20 log3(a5a6)^5=20 log3(a5a6)=4 a5a6=3^4=81

∵等比数列{an}中,an>0,a5a6=9,∴a1?a2…a10=(a5?a6)5=310,∴log3a1+log3a2+…+log3a10=log3310=10.故选B.

a5a6=a7a4=a8a3=…=a1a10,a1a2a3a4a5a6a7a8a9a10=(a5a6)的5次方=3的10次方,结果等于10,

LZ应该知道这条公式吧: lga+lgb=lg(ab) 所以你的问题不难,如果要分步做的话就是: 1、log3a1+log3a2=log3(a1*a2) 2、log3(a1*a2)+log3a3=log3[(a1*a2)*a3]=log3(a1*a2*a3) ..... 以此类推 得到log3a1+log3a2+log3a3+...+log3a10=log3(a1*...

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