作业帮化简：（log2(3)+log4(9)+log8(27)+…+log2n(3^n))*log2(8的n次方根)
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/03 20:01:57
化简：（log2(3)+log4(9)+log8(27)+…+log2n(3^n))*log2(8的n次方根)
化简：（log2(3)+log4(9)+log8(27)+…+log2n(3^n))*log2(8的n次方根)
化简：（log2(3)+log4(9)+log8(27)+…+log2n(3^n))*log2(8的n次方根)
A=m^k,B=n^k,则logA(B)=logm(n)
8^(1/n)=(2*2*2)^(1/n)=2^(3/n),故 log2(8^(1/n))=3/n
所以 log2(3)+log4(9)+log8(27)+…+log(2^n)(3^n)=n*log2(3)
所以 (log2(3)+log4(9)+log8(27)+…+log(2^n)(3^n))*log2(8^(1/n))=3*log2(3)
log2^x=log4^9
化简：（log2(3)+log4(9)+log8(27)+…+log2n(3^n))*log2(8的n次方根)
已知log2^3=a则log4^9=
log4分之9是否等于log2分之3
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算式log2^12-log4^9的值
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log2(20)-log2(5)+2log3(2)×log4(3)
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计算（log2 3+log4 9+log8 27+……+log2^n 3^n）log9 n次根号3计算（log2 3+log4 9+log8 27+……+log2^n 3^n）log9 n次根号32
log3^36-log3^4+log2(log2^16)+8^2/3+(27/8)^-1/3+log8^27/log4^9怎么算啊,不好意思哦,我已经没分啦.麻烦一下写一写log2(log2^16) (27/8)^-1/3 log8^27/log4^9
(log2 3+log4 9+log8 27+...+log2^n 3^n）log9 32的N次方根
(log2 3+log4 9+log8 27+...+log2^n 3^n）log9 32的N次方根
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[log3（4）+log根下3（2）][log2(9)+log4(根3)]
3^log9(25)+16^log4(2)-(根号7)^log7(9)+(-1)^log2(1)