log2(x-1)等于
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![log2(x-1)等于](/uploads/image/f/672072-24-2.jpg?t=log2%28x-1%29%E7%AD%89%E4%BA%8E)
若2*[log(1/2)x]的平方+7log(1/2)x+3小于等于0,求Y=log2(x/log1/2(x)=-log2(x)依题意有2(-log2(x))^2-7log2(x)+3
log2X是增函数所以只要比较真数的大小就行了但是还要考虑定义域2x+3>x+1移项X>-212X+3>0x>-3/22x+1>0x>-13i23求交集所以x>-1最后不要忘了写成集合的形式
底是2吧,x-1>0,x>0,log2(x-1)+log2x=log2(x-1)x
2根号2再问:只有1/3,√3/6,√3/3.√2/4这四个选项再问:里面那个错了,应该是log2x再答:应该是第四个再答:x是8再问:哦
log2(x²+1)
目测你搞错了,对数真数是大于0的,x大于0时-x-1必定小于0再问:那当x小于0时呢再问:当x小于0时y为什么等于log2(-x-1)打错了
答:|log2(x)-1|
log2(x-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
由x+4>0,x-1>0,x+8>0得x>1∵log2(x+4)+log2(x-1)=log2(2)+log2(x+8)∴log2(x+4)(x-1)=log2(x+8)*2,即(x+4)(x-1)=
f(2)=1+f(1/2)*log2^2=1+f(1/2)又f(1/2)=1+f(2)*log2^(1/2)=1-f(2)则f(2)=1,f(1/2)=0
(1,2)
log2(x+1)
再答:再答:再答:希望亲能看懂
log2(x+3)+log2(x+2)=1log2{(x+3)*(x+2)}=1(x+3)*(x+2)=2x^2+5x+4=0(x+4)(x+1)=0所以x=-4或-1因为x+3>0x+2>0所以x>
log2[x+1]≤log4[3x+1]定义:x+1>0x>-13x+1>0x>-1/3log2[x+1]≤(1/2)log2[3x+1]2log2[x+1]≤log2[3x+1]log2[x+1]&
f(x)=1+f(1/x)log2(x)(1)令把上面的x换为1/x,则f(1/x)=1+f(x)log2(1/x)(2)联立(1)和(2)式可得:f(x)=(1+log2(x))/(1+log2(x