题目内容
计算:
(1)2log52+log5
+loge
+3
×
×21-log23;
(2)log225•log34•log59.
(1)2log52+log5
| 5 |
| 4 |
| e |
| 1 |
| 2 |
|
(2)log225•log34•log59.
分析:(1)(2)根据对数运算法则和指数的运算法则进行计算;
解答:解:(1)2log52+log5
+loge
+3
×
×21-log23=log54+log5
+
logee+
×
×
=log5(4×
)+
+
×
=1+
+1
=
(2)log225•log34•log59
=2log25•2log32•2log53
=8×
×
×
=8
| 5 |
| 4 |
| e |
| 1 |
| 2 |
|
| 5 |
| 4 |
| 1 |
| 2 |
| 3 |
| ||
| 2 |
| 2 |
| 2log23 |
=log5(4×
| 5 |
| 4 |
| 1 |
| 2 |
| 3 |
| 2 |
| 2 |
| 3 |
=1+
| 1 |
| 2 |
=
| 5 |
| 2 |
(2)log225•log34•log59
=2log25•2log32•2log53
=8×
| lg5 |
| lg2 |
| lg2 |
| lg3 |
| lg3 |
| lg5 |
=8
点评:此题主要考查对数函数和指数函数的性质及其运算法则,计算的时候要仔细认真,是一道基础题;
练习册系列答案
相关题目