已知x^2+y^2=1,x>0,y>0,且loga(1+x)=m,loga(1/1-x)=n,则loga(y)=

解：∵2loga(y)=loga(y2)=log(1-x2)① ∵loga(1/1-x)=-loga(1-x)=n ∴loga(1-x)=-n② ∵loga(1+x)+loga(1-x)=loga[(1+x)(1-x)]=loga(1-x2)=m-n③ 联立①②③，得：loga(y)=(m-n)/2