早教吧 育儿知识 作业答案 考试题库 百科 知识分享

log2(x^2+1)+log2(y^2+4)=log2(8y)+log2(x),求log(y)x的值

题目详情
log2(x^2+1)+log2(y^2+4)=log2(8y)+log2(x),求log(y)x的值
▼优质解答
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