(lgx)^2-lgx^2-2=0,则(lgx)^2-2lgx-2=0,设y=lgx,则y²-2y-2=0,设两个根是y1,y2,
根据韦达定理,得y1+y2=2,y1*y2=-2
因为y1=lga,y2=lgb,则lga+lgb=2,lga*lgb=-2
loga b+logb a=lgb/lga+lga/lgb=lg²b/(lga*lgb)+lg²a/(lga*lgb)=(lg²a+lg²b)/(lga*lgb)
=[(lga+lgb)²-2lga*lgb]/(lga*lgb)=(2²+4)/(-2)=8/(-2)=-4