问题
解答题
求下列式子的值: (1)2log510+log50.25-log39; (2)0.25×(-
|
答案
(1)原式=lo
-log (102×0.25)5
=log 323
-2=0;g 525
(2)原式=0.25×16-4-3-2×(-
)1 2
=4-4-3=-3.
求下列式子的值: (1)2log510+log50.25-log39; (2)0.25×(-
|
(1)原式=lo
-log (102×0.25)5
=log 323
-2=0;g 525
(2)原式=0.25×16-4-3-2×(-
)1 2
=4-4-3=-3.