题目内容
(1)计算:0.064-1 |
3 |
1 |
8 |
3 |
4 |
1 |
2 |
(2)已知a=log32,3b=5,用a,b表示log3
30 |
分析:(1)利用根式与分数指数幂的运算法则,原式=((0.4)3)-
-1+23+((
)2)
=0.4-1 -1+8+
.
(2)由指数式和对数式的互化公式得b=log35,再利用对数式的运算性质,化简所求式子.
1 |
3 |
1 |
2 |
1 |
2 |
1 |
2 |
(2)由指数式和对数式的互化公式得b=log35,再利用对数式的运算性质,化简所求式子.
解答:解:(1)原式=((0.4)3)-
-1+23+((
)2)
,
=0.4-1 -1+8+
=
+7+
=10.
(2)∵3b =5,∴b=log35,
∴log3
=
log3(2×3×5)=
(log32+1+log35)=
(1+a+b).
1 |
3 |
1 |
2 |
1 |
2 |
=0.4-1 -1+8+
1 |
2 |
5 |
2 |
1 |
2 |
(2)∵3b =5,∴b=log35,
∴log3
30 |
1 |
2 |
1 |
2 |
1 |
2 |
点评:本题考查根式与分数指数幂的运算法则,指数式和对数式的互化,数式的运算性质.
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