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设x>3y>0,2log3(x-3y)=log3(3x+y)+log3y,求x/y的值
人气:409 ℃ 时间:2020-09-13 00:43:52
解答
2log3(x-3y)=log3(3x+y)+log3y
log3(x-3y)^2=log3[y(3x+y)]
(x-3y)^2=y(3x+y)
x^2-6xy+9y^2=3xy+y^2
x^2-9xy+8y^2=0
(x-y)(x-8y)=0
x=y,x=8y
x>3y>0
则x=y显然不成立
所以x=8y
x/y=8
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