log2(2^x-1)log1/2[2^(x+1)-2]0的取值范围？还有Y>1→x>log2(3)吧？我错了么？

解不等式log1/2(x+2)《2log2 1/x [log2(x)]^2+(a+1/a)log1/2(x)+1 Log1/2（x+1）+log2（1/（6-x））>log1/2(12) 解不等式log1/2^(x+1) +log2^[(1/(6-x)] 大于log1/2^12 8log2^X-logX^2=log1/2^X 解方程 log2(x-3)-log1/2(x)=2 log1/3 3 =2log2 x 求x 解不等式log2(x+1)+log1/4(x-1)〉log4(2x-1) 求函数f(x)=log2/5^(x+1)-log1/5^[(x+1)^2](0 解方程log2((2^-x)-1)*log1/2((2^-x+1)-2)=-2 求不等式log1/2(x+1)≥log2(2x+1)的解集 求函数y=log1/2(1-3^x)-log2(3^x+1/3)的最小值 解不等式log1/2(3-x)>=log2(x+3)-1 y=log2[1-x2] 和 y=log1/4(-2x2+x)的值域 已知x满足不等式log1/2（x）≥log1/2（2-x）,求函数f(x)=[log2(x/4)]*[log2(x/2)]的最小值 log1/2 (x^2)大于等于log1/2 (3x-2),求f(x)=(log2 (x/4))*(log2 (x/2))求最大值最小值 求B=｛X|log1/2（x^2+x+1）＞-log2（x^2+2）｝ log1/2(x^2+x+1)大于-log2(x^2+2),求X的取值范围