题目内容
计算:(1)(7 |
6 |
7 |
6 |
10 |
(2)log2(1+
2 |
3 |
2 |
3 |
分析:(1)利用
+
与
-
互为倒数,对数恒等式及对数的运算性质,运算求得结果.
(2)直接利用对数的运算性质,把要求的式子化为=log2(1+
+
)(1+
-
)=log2(2
)
=log22
,从而得到答案.
7 |
6 |
7 |
6 |
(2)直接利用对数的运算性质,把要求的式子化为=log2(1+
2 |
3 |
2 |
3 |
2 |
=log22
3 |
2 |
解答:解:(1)(
+
)2log(
-
)
=(
+
)log(
-
)10=(
-
)-log(
-
)10
=(
-
)log(
-
)
=
,故答案为:
.
(2)log2(1+
+
)+log2(1+
-
)=log2(1+
+
)(1+
-
)=log2(2
)
=log22
=
,
故答案为:
.
7 |
6 |
7 |
6 |
10 |
7 |
6 |
7 |
6 |
7 |
6 |
7 |
6 |
=(
7 |
6 |
7 |
6 |
1 |
10 |
1 |
10 |
1 |
10 |
(2)log2(1+
2 |
3 |
2 |
3 |
2 |
3 |
2 |
3 |
2 |
=log22
3 |
2 |
3 |
2 |
故答案为:
3 |
2 |
点评:本题考查对数恒等式,对数的运算性质的应用,式子的变形是解题的难点.
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