请问log2 (3x/x+1)=-1该怎么解?

请问log2 (3x/x+1)=-1该怎么解? log2 (x + 3) + log2(x + 2) = 1log2 (x + 3) + log2(x + 2) = 1 |[log2(x)]^2-3log2(x)+1| log2(x+1)>log2(3-x) log2 (x+1)+log2 x=log6 求解log2(3x)=log2(2x+1) log2(x+1)+log2(x+8)=3的解集是? log2(x)=-1 请问y=log2[x+根号(1+x^2)] 奇偶性 [log2 1]+[log2 2]+[log2 3]+[log2 4]+[log2 5]+...+[log2 1024]=?[x]表示不超过x的最大整数2为底 答案是8204 log2(4^x+4)=x+log2(2^x+1-3)log2(4^x+4)=x+log2[2^(x+1)-3] 解方程log2(2-x)=log2(x-1)+1 解2log2^(x-5)=log2^(x-1)+1 不等式log2(2x+3)>log2(x+1)的解集是 y=log2^(x+1) 3 log2(x+1)=3怎么解 解方程log2(4^x+1)=x+log2(2^(x+3)-6) 解方程log2^(4^x+4)=x+log2^[2^(x+1)-3]