题目内容

(1)若log23=a,3b=7求用a、b表示log4256
(2)logac,logbc是方程x2-3x+1=0的两根,求log
ab
c
的值
分析:(1)由题意知log37=b,log23•log37=log27=ab.由此可知log4256的值.
(2)由题意知logac•logbc=1,logca+logcb=
logac+logbc
logac•logab
=3,log
a
b
c
=
1
logca-logcb
=
1
(logca+logcb) 2-4logca•logcb
,代入可得答案.
解答:解:(1)∵3b=7,∴log37=b,
∴log23•log37=log27=ab.
∴log4256=
log256
log242
=
log27+3log22
log22+log23+log27

=
ab+3
1+a+ab

(2)logac+logbc=3,logac•logbc=1,
logca+logcb=
logac+logbc
logac•logab
=3,
log
a
b
c
=
1
logca-logcb
=
1
(logca+logcb) 2-4logca•logcb

=
1
9-4
=
5
5
点评:本题考查对数的性质和运算法则,解题时要认真审题,仔细解答.
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