log2（2x-1）＞log1/2（1-3x）

3个回答

由log1/2(1-3x)=-log2(1-3x)可知：
移项得log2(2x-1)+log2(1-3x)>0
即log2((2x-1)*(1-3x))=log2(5x-6x^2-1)>0
即5x-6x^2-1>1
6x^2-5x+21!

log2(x+1)>log2(3-x)
3+log1/2x=log2x,x=?
已知log2[log1/2(log2x)]=log3[log1/3(log3y)]=log5[log1/5(log5z)
log 9 X/Y=1/2(log 3 X - log 3 Y) log 9 X/Y=1/4 3log 3 X =2lo
解方程式 log2(x+3)-log2(x-1)＝1
不等式log1/3(x^2-8)>log3(1/2x)
log2x^2=log2(根号2-1)-log2(根号2+1),求x+1/x
log2(2x-1)2+log2x=1
已知-3≤log1/2 x≤-1,f(x)=[log2(x*4^m)]*[log2(4/x)]
设函数f(x)＝log2x+1x−1+log2(x−1)+log2(p−x)，