Question

高一数学一道题~~~~~20分在线等待

计算。
1.log2 3+log4 9+log8 27+log9 √32
注意；√表示根号哈。log2 3,2是底数。3是真数。
麻烦大家把过程写清楚哈，谢啦。

Answer 1

[1]
利用换底公式及对数性质.可知
log2,3=(ln3)/(ln2)
log4,9=(ln9)/(ln4)=(ln3²)/(ln2²)=(2ln3)/(2ln2)=(ln3)/(ln2)
log8,27=(ln27)/(ln8)=(ln3³)/(ln2³)=(3ln3)/(3ln2)=(ln3)/(ln2)
[2]
log9, √32=(ln√32)/(ln9)=[(5/2)ln2]/(2ln3)=[(5/4)ln2]/(ln3)
[3]
原式=[(3ln3)/(ln2)]+[(5ln2)/(4ln3)]
=[3log2, 3]+[(5/4)log3, 2]

Answer 2

log2 3+log4 9+log8 27+log9 √32
=lg3/lg2+lg9/lg4+lg27/lg8+lg32/lg9
=lg3/lg2+2lg3/(2lg2)+3lg3/(3lg2)+5lg2/(2lg3)
=lg3/lg2+lg3/lg2+lg3/lg2+5lg2/(2lg3)
=3lg3/lg2+5lg2/(2lg3)
=3lg3/lg2+5/2×lg2/lg3

希望对您有所帮助
如有问题，可以追问。
谢谢您的采纳

Answer 3

Log(2,3)+log(4,9)+log(8,27)+log(9,√32)
=ln3/ln2+ ln3/ln2+ ln3/ln2+5ln2/(4ln3)
=3 ln3/ln2+5ln2/(4ln3)
=3 ln3/ln2+5/(4ln3/ln2)
=[12( ln3/ln2)^2+5]/(4ln3/ln2)
=[12(log(2,3))^2+5]/(4log(2,3))

Answer 4

原式=log2 3 + log(2^2) (3^2) + log(2^3) (3^3) + log(3^2) [32^(1/2)]
=log2 3 + (2/2)log2 3 + (3/3)log2 3 + [(1/2)/2]log3 32
=3log2 3 + (1/4)log3 (2^5)
=3log2 3 + (5/4)log3 2

Answer 5

(log2 3) + (log2 9)/(log2 4) + (log2 27)/(log2 8) + (log2 √32)/(log2 9)
(log2 3) + (log2 3^2)/(log2 2^2) + (log2 3^3)/(log2 2^3) + (log2 √2^5)/(log2 3^2)
(log2 3) + (log2 3) + (log2 3) + (log2 2^(5/2))/(log2 3^2)
3*(log2 3) + (5/2)/(log2 3^2)
3*(log2 3) + (5/4)/(log2 3)
3*(log2 3) + (5/4)(log3 2)

Answer 6

log4 9=log2^2 3^2=log2 3
log8 27=log2^3 3^3=log2 3
log9 √32=log3^2 2^(5/2)=5/4*log2 3
结果是17/4*log2 3