log2 1+ log2 2+ … + log2 n >= ⌊n/2」log2 (n/2) 求证明,2是底

log2*(2分之1)^(n-1))=1-n求计算过程 (log2 9)/(log2 8) = 2/3(log2 3) log2 SQR(7/48)+log2 12 -1/2log2 48 log2 SQR(7/48)+log2 12 -1/2log2 42 在等比数列{an}中,a1=2,a4=16,令bn=1/｛log2(an).log2[a(n+1)]｝ 已知log2（2m-4）+log2（n-4）=3，则m+n的最小值为 ___ ． 2^(log2^(-3))= 解方程log2(2-x)=log2(x-1)+1 log2 (x + 3) + log2(x + 2) = 1 求函数y=log2(1+sinx)+log2(1-sinx)[注log2(1+sinx)是log以2为底的对数函数】 Tn=log2[2^1]+log2[2^(-2)]+...+log[2^(4-3n)] =1-2-5+.+4-3n =( log2底√1/48+log2底12-1/2log2底24等于?