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log2(x^2+1)+log2(y^2+4)=log2(8y)+log2(x),求log(y)x的值
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log2(x^2+1)+log2(y^2+4)=log2(8y)+log2(x),求log(y)x的值
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答案和解析
log2(x^2+1)+log2(y^2+4)=log2(8y)+log2(x),
(x^2+1)(y^2+4)=8yx
x^2y^2+4x^2+y^2+4-8xy=0
(xy-2)^2+(2x-y)^2=0
xy=2,2x-y=0
xy=2,2x=y
2x^2=2
x=±1
又由log2(x),可知x>0
即x=1,y=2
log(y)x=log(2)1=0
(x^2+1)(y^2+4)=8yx
x^2y^2+4x^2+y^2+4-8xy=0
(xy-2)^2+(2x-y)^2=0
xy=2,2x-y=0
xy=2,2x=y
2x^2=2
x=±1
又由log2(x),可知x>0
即x=1,y=2
log(y)x=log(2)1=0
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