Question

如果方程lg²x-lgx²-2=0有两根a,b那么loga b+logb a的值等于

Answer 1

a,b是方程（lgx）^2-lgx^2-2=0的两个根

即:

(lga)^2-2lga-2=0

(lgb)^2-2lgb-2=0

设m=lga,n=lgb

则:m^2-2m-2=0,n^2-2n-2=0

说明m,n是方程t^2-2t-2=0的二个根.

m+n=2,mn=-2

loga b+logb a=lgb/lga+lga/lgb=[(lgb)^2+(lga)^2]/(lga*lgb)

=[(m+n)^2-2mn]/(mn)

=[4-2*(-2)]/(-2)

=-4