3+log2-7/1+log2-7=3+log3-7×log2-3/1+log3-7×log23 前者是怎么变成后者的?（ps：我为了区分底数和真数,我在底数和真数的中间划了"-"做标记.还有“/"其实是分数形式.）最后的log23 应该是log2-3 打错了

log2(3)×log2(4)=log2(7)?log2(3)×log2(4)=log2(7)吗?log2(1/49)+log2(1/16)+log2(1/27)等于多少呢?帮个忙! 求值：log2 (5/56)^(1/2)+log2 (14)-(1/2)log2 (35)=log2中2为底数=1/2[-2log2 (7)-3]+log2 (2)+log2 (7) 中的1/2[-2log2 (7)-3]是怎么得到的？ log2^(4^7*2^5)+log2^6-log2^3 log2 3×log3 7=log2 7这个算法对吗 化简算式log2^√7/48+log2^12-1/2log2^42+1/2^log2^3 log2 14-2log2 7/3+log2 7-log2 18是多少2是小2 log2 (x + 3) + log2(x + 2) = 1log2 (x + 3) + log2(x + 2) = 1 方程log2(9^(x-1)+7)=2+log2(3^(x-1)+1)的解? 方程log2[9^(x-2) +7]=2+log2[3^(x-1) +1]的解为 log2 6-log2 3 log2^6-log2^3 log2(4)/log2(3) [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 SQR(7/48)+log2 12 -1/2log2 42 log2 SQR(7/48)+log2 12 -1/2log2 48 (log2 9)/(log2 8) = 2/3(log2 3) 已知log2 m=-1/log2 3,则m= 求解log2(3x)=log2(2x+1)