题目内容
计算:
(I)
-(π-1)0-(3
)
+(
)-
;
(II)log2(47×25)+log26-log23.
(I)
6
|
3 |
8 |
1 |
3 |
1 |
64 |
2 |
3 |
(II)log2(47×25)+log26-log23.
分析:(I)根据
=
,a0=1(a≠0),a
=
,可计算出(I)式的值;
(II)根据logaM+logaN=loga(M•N),logaM-logaN=loga(M÷N),logaNm=mlogaN,可计算出(II)式的值;
n | an |
|
n |
m |
m | an |
(II)根据logaM+logaN=loga(M•N),logaM-logaN=loga(M÷N),logaNm=mlogaN,可计算出(II)式的值;
解答:解:(I)
-(π-1)0-(3
)
+(
)-
=
-1-
+
=
-1-
+16
=16
(II)log2(47×25)+log26-log23
=log2(47×25×6÷3)
=log2(214×25×2)
=log2(220)
=20
6
|
3 |
8 |
1 |
3 |
1 |
64 |
2 |
3 |
=
|
3 |
| ||
3 | (43)2 |
=
5 |
2 |
3 |
2 |
=16
(II)log2(47×25)+log26-log23
=log2(47×25×6÷3)
=log2(214×25×2)
=log2(220)
=20
点评:本题考查的知识点是有理数指数幂的化简求值,对数的运算性质,熟练掌握指数和对数的运算性质是解答的关键.
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