Question

一道高考的数学题谢谢给讲一下~~

{{[(1-log6(3)]}²+log6(2)×log6(18)}÷log6(4)

已知log18(9)=a,log18(5)=b,求log36(15)
给讲一下，谢谢

Answer 1

记:以a为底b的对数为loga(b).

由1-log6(3)
=log6(6)-log6(3)
=log6(6/3)
=log6(2).

又log6(18)

=log6(3×6)
=log6(3)+log6(6)
=1+log6(3).

从而log6(2)×log6(18)

=log6(2)×(1+log6(3).

log6(4)

=log6(2²)

=2log6(2).

{{[(1-log6(3)]}²+log6(2)×log6(18)}÷log6(4)

={[log6(2)]²+[log6(2)×(1+log6(3)]}÷[2log6(2)]
={log6(2)+[1+log6(3)]}÷2

={1+log6(2×3)}÷2
={1+log6(6)}÷2

=(1+1)÷2
=1.

由
已知log18(9)=a，

得2log18(3)=a
从而log18(3)=a/2.

而log18(2)=log18(18/9)
=log18(18)-log18(9)
=1-log18(9)
=1-a.
又log18(5)=b.
由换底公式有 ：
log36(15)

=log18(15)/log18(36)
=[log18(3)+log18(5)]/[2log18(6)]
=[log18(3)+log18(5)]/[2log18(2)+2log18(3)]

=[(a/2)+b]/[2(1-a)+2a/2]
=[(a+2b)/2]/(2-a)
=(a+2b)/[2(2-a)]

=

Answer 2

解答：
(1)
{{[(1-log6(3)]}²+log6(2)×log6(18)}÷log6(4)

={{[(log6(6)-log6(3)]}²+log6(2)×log6(18)}÷log6(4)
={[log6(2)]²+log6(2)×log6(18)}÷log6(4)
=log6(2)*[log6(2)+log6(18)]÷log6(4)
=log6(2)*log6(36)÷log6(4)
=2log6(2)÷log6(4)
=1

(2)

log18 (9)=a,

18^b=5----->log18 (5)=b
log36(15)
=log18(15) /log18(36)
=[log18(5)+log18 (3)]/[log18(18)+log18(18)-log18(9)]
=(b+a/2)/(2-a)
=(2b+a)/(4-2a)

Answer 3

啊啊，解不清楚

Answer 4

题设不清

Answer 5

我不会