log2底x=2^x

log2底x=2^x log2 (x + 3) + log2(x + 2) = 1log2 (x + 3) + log2(x + 2) = 1 log2^(x^log2^x)=? [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(2x)*log2(x)=2 x=? f(x)=log2(x/8)*log2(2/x) log2(x – 2 ) + 2 = log2(5)log2(x – 2 ) + 2 = log2(5) log2(x – 2 ) + 2 = log2(5)log2(x – 2 ) + 2 = log2(5) log2 4+log2 (x+2)怎么算?log2 4+log2 (x+2)=log2 (4x+8) log2(-x)=x+ |2x-log2^x| 求导 y=log2(x^2)-log2(x) 求解log2(3x)=log2(2x+1) 解方程log2(2-x)=log2(x-1)+1 解2log2^(x-5)=log2^(x-1)+1 解方程log2(x+14)-log2(x+6)=3-log2(x+2) log4(log2^x)=log2(log4^x)求log2^x log2 (x+1)+log2 x=log6