题目内容
设logac,logbc是方程x2-3x+1=0的两个根,求log
c的值.
| a |
| b |
考点:对数的运算性质
专题:计算题
分析:由方程x2-3x+1=0解得x=
.由于logac,logbc是方程x2-3x+1=0的两个根,可得logac=
,logbc=
;或logac=
,logbc=
.可得lga-lgb=±
lgc,再利用对数的换底公式可得log
c=
即可得出.
3±
| ||
| 2 |
3+
| ||
| 2 |
3-
| ||
| 2 |
3-
| ||
| 2 |
3+
| ||
| 2 |
| 5 |
| a |
| b |
| lgc |
| lga-lgb |
解答:
解:由方程x2-3x+1=0解得x=
.
∵logac,logbc是方程x2-3x+1=0的两个根,
∴logac=
,logbc=
;或logac=
,logbc=
.
①当logac=
,logbc=
时,
∴
=
,
=
.
∴lga-lgb=-
lgc,
∴log
c=
=
=-
;
②当logac=
,logbc=
时.
lga-lgb=
lgc,
∴log
c=
=
.
3±
| ||
| 2 |
∵logac,logbc是方程x2-3x+1=0的两个根,
∴logac=
3+
| ||
| 2 |
3-
| ||
| 2 |
3-
| ||
| 2 |
3+
| ||
| 2 |
①当logac=
3+
| ||
| 2 |
3-
| ||
| 2 |
∴
| lgc |
| lga |
3+
| ||
| 2 |
| lgc |
| lgb |
3-
| ||
| 2 |
∴lga-lgb=-
| 5 |
∴log
| a |
| b |
| lgc |
| lga-lgb |
| lgc | ||
-
|
| ||
| 5 |
②当logac=
3-
| ||
| 2 |
3+
| ||
| 2 |
lga-lgb=
| 5 |
∴log
| a |
| b |
| lgc | ||
|
| ||
| 5 |
点评:本题考查了对数的运算法则和对数的换底公式,属于中档题.
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