题目内容
计算下列各式.
(1)
lg25+lg2-lg
-log29×log32;
(2)64
-(-
)0+(
)-
+lg20+log10025.
(1)
| 1 |
| 2 |
| 0.1 |
(2)64
| 1 |
| 3 |
| 2 |
| 3 |
| 1 |
| 16 |
| 1 |
| 2 |
分析:(1)利用对数的运算法则即可得出;
(2)利用指数幂和对数的运算法则即可得出.
(2)利用指数幂和对数的运算法则即可得出.
解答:解:(1)原式=
lg52+lg2-
lg10-1-
×
=lg5+lg2+
-2
=1-
=-
.
(2)原式=43×
-1+2-4×(-
)+lg20+
=4-1+22+lg20+lg5
=7+lg(20×5)
=7+2=9.
| 1 |
| 2 |
| 1 |
| 2 |
| 2lg3 |
| lg2 |
| lg2 |
| lg3 |
=lg5+lg2+
| 1 |
| 2 |
=1-
| 3 |
| 2 |
=-
| 1 |
| 2 |
(2)原式=43×
| 1 |
| 3 |
| 1 |
| 2 |
| lg52 |
| lg102 |
=4-1+22+lg20+lg5
=7+lg(20×5)
=7+2=9.
点评:本题考查了指数幂和对数的运算法则,属于基础题.
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