已知x^2+y^2=1,x大于0,y大于0,且loga(1+x)=m,loga(1/1-x)=n,则loga(y)为多少?
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已知x^2+y^2=1,x大于0,y大于0,且loga(1+x)=m,loga(1/1-x)=n,则loga(y)为多少?
已知x^2+y^2=1,x大于0,y大于0,且loga(1+x)=m,loga(1/1-x)=n,则loga(y)为多少?
已知x^2+y^2=1,x大于0,y大于0,且loga(1+x)=m,loga(1/1-x)=n,则loga(y)为多少?
x^2+y^2=1
y=根号(1-x^2)=根号(1-x)(1+x),
所以loga(y)=loga根号(1-x)(1+x)
=1/2loga(1-x)(1+x)
=1/2[loga(1-x)+loga(1+x)]
=1/2[-loga(1/1-x)+loga(1+x)]
=1/2(-n+m)
=1/2(m-n)
因为x^2+y^2=1,x大于0,y大于0,
所以y=根号(1-x^2)=根号(1-x)(1+x),
所以loga(y)=loga根号(1-x)(1+x)
=1/2loga(1-x)(1+x)
=1/2[loga(1-x)+loga(1+x)]
=1/2[-loga(1/1-x)+loga(1+x)]
=1/2(-n+m)
=1/2(m-n)