作业帮log2(x-1)+log2(x+1)=3的解是?
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log2(x-1)+log2(x+1)=3的解是?
log2(x-1)+log2(x+1)=3的解是?
log2(x-1)+log2(x+1)=3的解是?
log2(x-1)+log2(x+1)=3
log2(x-1)(x+1)=3
函数定义域为
(x-1)(x+1)>0
所以x>1或x
log2x-2+log2x+2=3
4x-2+2=3
4x=3
x=0.75(3/4)
log2(x-1)+log2(x+1)=3
log2(x-1)(x+1)=3
(x-1)(x+1)=2^3
x^2-1=8
x^2=9
x=3(另一根x=-3不合题意要舍去）
X=3
条件log2(x-1)+log2(x+1)=3
得x-1>0
x+1>0
(x-1)(x+1)=2^3=8
得x=3(x=-3不符合x>1,所以舍去）
log2 (x+1)+log2 x=log6
log2(x)=-1
log2 (x + 3) + log2(x + 2) = 1log2 (x + 3) + log2(x + 2) = 1
|[log2(x)]^2-3log2(x)+1|
log2(x+1)>log2(3-x)
解方程log2(2-x)=log2(x-1)+1
解2log2^(x-5)=log2^(x-1)+1
求解log2(3x)=log2(2x+1)
log2(x+1)+log2(x+8)=3的解集是?
log2^(x^log2^x)=?
log2(log5 x)=-1
谁能帮我解一下：log2(x + 1) = 4 - log2(x + 1)log2(x + 1) = 4 - log2(x + 1)
解方程：log2(x+4)+log2(x-1)=1+log2(x+8)
不等式log2 (x+1)
log2(2X-1)
log2(x²+1)
log2(x-1)
log2(x+1)