若方程2log2x-log2(x-1)=m 1

log2(x+2)-log2x=log2(1+2/x)=a实数a的取值范围(log2(5/3),log2(3/2))

（log2x)²-log2(x³)+2=0（log2x)²-3log2(x)+2=0(log2(x)-1)(log2(x)-2)=0log2(x)=1或2x=2或4

底是2吧,x-1>0,x>0,log2(x-1)+log2x=log2(x-1)x

解析：由题意可知：x>0,则原不等式可化为：(log2x)^2-2log2x≤0(log2x)[(log2x)-2]≤0解得：0≤log2x≤2即得：1≤x≤4所以原不等式的解集为{x|1≤x≤4}

log2(x+3)/x

Y=log2(x^2+1)-log2x零和负数无对数：x＞0Y=log2(x^2+1)-log2x=log2{(x^2+1)/x}=log2{x+1/x}x＞0,x+1/x≥2,log2{x+1/x}

log1/2(log2x)=1--->log2x=1/2-->x=2^(1/2)log1/3(log3y)=1--->log3y=1/3-->y=3^(1/3)x^6=8再问：x^6=8

f(x)=(log2(x)-1)(log2(x)-2)=[log2(x)-3/2]^2-/4当log2(x)=3/2时取得最小值=-1/4由于f(x)的对称轴x=3/23/2-1

log2(2-x)=log2(x-1)+log(2)log2(2-x)=log2[2(x-1)]2-x=2x-2x=4/3带入检验,成立所以x=4/3

log2(x+14)-log2(x+6)=3-log2(x+2)∴log2[(x+14)/(x+6)]=log2[8/(x+2)]∴(x+14)/(x+6)=8/(x+2)且(x+14)/(x+6)>

log2(x/2)〕×〔log√2(√x/2)〕=[(log2X)-1]*[(log2X)-2]=[(log2X)-3/2]²-1/4实数x满足不等式1/2≤log2X≤3-1/4≤[(lo

log2(x²)=log2[(√2-1)/(√2+1)]log2(x²)=log2(√2-1)²所以x²=(√2-1)²x=±(√2-1)x=√2-1

第一个不等式：0

（1）1/4≤x≤4log2(1/4)=12log2(4)=2所以t∈（-2,2）（2）f(x)=(log4+logx)(log2+logx)(底数2我就不写了~)令t=logx则f(x)=(t+2)

∵t＝㏒2x在定义域x＞0上是增函数1/4≤x≤4∴㏒2(1/4)≤t≤㏒24∴﹣2≤t≤2f(x)＝㏒2(4x)×㏒2(2x)＝(2＋㏒2x)×(1＋㏒2x)＝(2＋t)(1＋t)＝t²＋

∵log2x+log2（x-1）=1，∴log2（x2-x）=1，即x2-x=2，解得x=-1，或x=2，又∵x＞0，x-1＞0，∴函数的定义域是x＞1，M={2}；若22x+1-9•2x+4=0，∴

/>f(x)=log2x/2*log2x/4=(log2x-1)(log2x-2)=(log2x)^2-3log2x+2=(log2x-3/2)^2-1/4∵1≦log2x≦2∴当log2x=3/2时

∵log2[log12(log2x)]＝log3[log13(log3y)]＝log5[log15(log5z)]=0，∴log12(log2x) ＝log13(log3y)=log15(l

令X=log2x,则X^2+(t-2)X+1-t-m=0m=X^2+(t-2)X+1-t=(X+(t-2)/2)^2-t^2/4t∈[-2,2]0≤t^2/4≤1显然t^2/4＝0,X≠1当t=2时,

我们的公式是log2（4x）+log2（2x）=log2（2x*4x）没有log2（4x）log2（2x）=log2（2x+4x）这个公式再问：log2（4x）*log2（2x）=log2(4x+2x