分类：教育/科学-理工学科-数学
问题：不等式已知x满足2(log1/2x)^2+7log1/2x+3≤ 
已知x满足2(log1/2 x)^2+7log1/2 x+3≤0,求f(x)=  ( log1/2 x/2)*(log2 x/4)的最值。         注：  (log1/2 x)这里表示以1/2为底的x
回答：2(logx)^2+7logx+3=<0 本题中省略底数1/2.
--->(2logx+1)(logx+3)=<0
--->-3=<logx=<-1/2.
f(x)=log(x/2)log2(x/4)
=log(x/2)[log(x/4)/log(1/2)]
=-(logx+1)(logx+2)  
 [注：log2 (x/2)=log(x/4)/log2=(logx-log4)/(-1)=-[logx-(-2)]=-(logx+2)]
=-(logx)^2-3logx-2
=-(logx+3/2)^2+1/4
-3=<logx=<-1/2--->-3/2=<logx+3/2=<1
--->0=<(logx+3/2)^2=<9/4
--->-9/4=<-(logx+3/2)^2=<0
--->-2=<(logx+3/2)^2=<1/4
所以f(x)min=-2,(logx=-3/2--->x=2^(3/2)=22)
f(x)max=1/4.(logx=-3--->x=2^3=8)