- 指数与指数幂的运算
- 共1477题
计算下列各式的值
(1)0.064 -13-(-)0+160.75+0.25 12
(2)lg5+(log32)•(log89)+lg2.
正确答案
(1)0.064 -13-(-)0+160.75+0.25 12
=((0.4)3)-13-1+(24)34+(0.52)12
=(0.4)-1-1+8+0.5
=2.5-1+8+0.5
=10;
(2)lg5+(log32)•(log89)+lg2
=lg5+•
+lg2
=1+•
=1+=
.
(1)计算:0.25-2+()-13-
lg16-2lg5+(
1
3
)0;
(2)解方程:log2(9x-5)=log2(3x-2)+2.
正确答案
(1)0.25-2+()-13-
lg16-2lg5+(
1
3
)0
=16+-lg4-lg25+1
=16+-2+1
=.
(2)∵log2(9x-5)=log2(3x-2)+2,
∴log2(9x-5)=log24(3x-2)
则原方程等价于,
∴(3x)2-4•3x+3=0,即(3x-3)(3x-1)=0,
∵3x>2,∴3x=3,∴x=1.
经检验,得原方程的根为x=1.
计算:
(1)(2)0+2-2×(2
)-12-(0.01)0.5;
(2)lg14-21g+lg7-lg18.
正确答案
(1)(2)0+2-2×(2
)-12-(0.01)0.5
=1+×
-0.1
=1+-
=.
(2)lg14-21g+lg7-lg18
=lg(14÷×7÷18)
=lg1
=0.
化简求值
(1)若x>0,化简 (2x 14+3 32)(2x 14-3 32)-4x -12(x-x 12).
(2)计算:2(lg)2+lg
•lg5+
.
正确答案
解析:(1)原式=(2x14)2-(332)2-4x1-12+4x-12+12=4x12-27-4x12+4=-23.
(2)原式=lg(2lg
+lg 5)+
=lg(lg 2+lg 5)+|lg
-1|
=lg+(1-lg
)=1.
(1)计算log3+lg25+lg4+7log72
(2)已知x12+x-12=3,求的值.
正确答案
解(1)log3+lg25+lg4+7log72
=log3+lg52+lg22+2
=-+2(lg5+lg2)+2
=;
(2)由x12+x-12=3,
得:(x12+x-12)2=9,
所以,x+2+x-1=9,
故x+x-1=7,
所以,=
=
.
扫码查看完整答案与解析