log(1/3)^x+log3^(1/3-x)
=log(1/3)^x+log(1/3)^[1/(1/3-x)]=
=log(1/3)^[x/(1/3-x)]=>=log(1/3)^(1/3)
所以x>0,且1/3-x>0,x/(1/3-x)<=1/3
{x|0
|LOG1/3X|(其中1/3是底数,X是真数)+|LOG3(1/3-X)|(其中3是底数,1/3-X是真数)>=1
|LOG1/3X|(其中1/3是底数,X是真数)+|LOG3(1/3-X)|(其中3是底数,1/3-X是真数)>=1
数学人气：448 ℃时间：2019-08-20 16:48:27
优质解答
log(1/3)^x+log3^(1/3-x)>=1,
log(1/3)^x+log3^(1/3-x)
=log(1/3)^x+log(1/3)^[1/(1/3-x)]=
=log(1/3)^[x/(1/3-x)]=>=log(1/3)^(1/3)
所以x>0,且1/3-x>0,x/(1/3-x)<=1/3
{x|0
log(1/3)^x+log3^(1/3-x)
=log(1/3)^x+log(1/3)^[1/(1/3-x)]=
=log(1/3)^[x/(1/3-x)]=>=log(1/3)^(1/3)
所以x>0,且1/3-x>0,x/(1/3-x)<=1/3
{x|0
我来回答
类似推荐