【lg^2*5+lg21g50log3*根号3/3+log8*44^log2*5+2*5^log25*4】
更新时间： 2025-08-27 07:19:16
1人问答
问题描述：
lg^2*5+lg21g50log3*根号3/3+log8*44^log2*5+2*5^log25*4
汪家铭回答：
(lg5)^2+(lg2)(lg50)
=(lg5)^2+(lg2)(lg25×2)
=(lg5)^2+(lg2)[lg(5^2)+lg2]
=(lg5)^2+(lg2)(2lg5+lg2)
=(lg5)^2+2(lg5)(lg2)+(lg2)^2
=(lg5+lg2)^2
=[lg(5×2)]^2
=(lg10)^2
=1^2
=1
log3*(√3/3)+log8*(4)
=[lg(√3/3)]/lg3+lg4/lg8
=(lg√3-lg3)/lg3+(lg2^2)/(lg2^3)
=[lg3^(1/2)-lg3]/lg3+(2lg2)/(3lg2)
=[(1/2)lg3-lg3]/lg3+2/3
=(1/2)-1+2/3
=2/3-1/2
=1/6
4^[log2*(5)]+2×5^[log25*(4)]
=(2^2)^[log2*(5)]+2×[25^(1/2)]^[log25*(4)]
=2^2[log2*(5)]+2×25^(1/2)[log25*(4)]
={2^[log2*(5)]}^2+2×{25^[log25*(4)]}^(1/2)
=5^2+2×4^(1/2)
=25+2×2
=29
写得够清楚了吧同学,因为我实在是无聊,已经很晚了……
数学推荐
![【(1-tanα)/(1+tanα)=3-2√2求[（sinα+cosα)^2-1]/(cotα-sinα乘cosα)的值】](/UploadFile/wdimg/2007/122438408/165/1665101409.jpg)
数学推荐
最新更新
精品分类
优秀数学推荐
热门数学