log2(8-2x-x^2)≤3 ~~~~~~~~~~~~~~~~~~~~~~要求详细过程

来源：学生作业帮助网 编辑：六六作业网 时间：2024/12/25 21:25:51
log2(8-2x-x^2)≤3~~~~~~~~~~~~~~~~~~~~~~要求详细过程log2(8-2x-x^2)≤3~~~~~~~~~~~~~~~~~~~~~~要求详细过程log2(8-2x-x^

log2(8-2x-x^2)≤3 ~~~~~~~~~~~~~~~~~~~~~~要求详细过程
log2(8-2x-x^2)≤3 ~~~~~~~~~~~~~~~~~~~~~~
要求详细过程

log2(8-2x-x^2)≤3 ~~~~~~~~~~~~~~~~~~~~~~要求详细过程
3=log2(8)
底数2大于1,log2(x)是增函数
所以8-2x-x^2=0
x(x+2)>=0
x=0
真数大于0
8-2x-x^2〉0
x^2+2x-8

8-2x-x^2≤2^3=8
x^2+2x>=0
x(x+2)>=0
x>=0或X<=-2

将式子化为8-2x-x^2≤2^3，则可化为-2x-x^2≤0，求得x>=0或者x<=-2。
并且8-2x-x^2>0故求得:-4综上所述可知0<=x<2或-4

8-2x-x^2<=8,8-2x-x^2>0
x^2+2x>=0
x>=0或x<=-2
8-2x-x^2>0
x^2+2x-8<0
-4所以-4

log2(8-2x-x^2)≤3=log2(2^3)
所以有:
0<8-2x-x^2<=8
(i)8-2x-x^2>0
x^2+2x-8<0
(x+4)(x-2)<0
解得:-4(ii)8-2x-x^2<=8
x^2+2x>=0
x(x+2)>=0
解得:x>=0或x<=-2
综上所述,0<=x<2或-4

8-2x-x^2<=9
x^2+2x+1>=0
(x+1)^2>=0
所以x任意值
但是
8-2x-x^2>0
-4

0<8-2X-X^2<=8
解这个不等式,可得x属于[-4,-2)并(0,2]

|[log2(x)]^2-3log2(x)+1| log2 (x + 3) + log2(x + 2) = 1log2 (x + 3) + log2(x + 2) = 1 f(x)=log2(x/8)*log2(2/x) |2x-log2^x| log2 4+log2 (x+2)怎么算?log2 4+log2 (x+2)=log2 (4x+8) 解不等式log2(4x+8)>log2(2x+1) 求函数Log2(8x).Log2(4x)其中2 求函数Log2(8x).Log2(4x)其中2 求解log2(3x)=log2(2x+1) 不等式log2(2x+3)>log2(x+1)的解集是 化简：(log2(x/4))*(log2(x/2)) log2（3－x）≤2 不等式log2(2x-3) log2(x^2-3)>0 解方程log2(x+14)-log2(x+6)=3-log2(x+2) log2(x)=-2/3,求x log2(2X-1) log2(x^2)图像