题目内容
计算下列各式的值.
(1)log2
+log212-
log242;
(2)lg52+
lg8+lg5•lg20+lg22.
(1)log2
|
| 1 |
| 2 |
(2)lg52+
| 2 |
| 3 |
考点:对数的运算性质
专题:函数的性质及应用
分析:利用对数的运算法则和运算性质求解.
解答:
解:(1)log2
+log212-
log242
=log2(
×12×
)
=log2(
)=log22-
=-
.
(2)lg52+
lg8+lg5•lg20+lg22
=2lg5+2lg2+lg5(1+lg2)+lg22
=2(lg5+lg2)+lg5+lg2(lg2+lg5)
=2+lg5+lg2=3.
|
| 1 |
| 2 |
=log2(
| ||
4
|
| 1 | ||||
|
=log2(
| 1 | ||
|
| 1 |
| 2 |
| 1 |
| 2 |
(2)lg52+
| 2 |
| 3 |
=2lg5+2lg2+lg5(1+lg2)+lg22
=2(lg5+lg2)+lg5+lg2(lg2+lg5)
=2+lg5+lg2=3.
点评:本题考查对数式的求解,是基础题,解题时要注意对数的运算法则和运算性质的合理运用.
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