求loga^b+logb^a的值已知a,b是方程(log2^x)^2-log2^x^2-2=0的两个根,求loga^b+logb^a的值.--------------我知道是通过两根之和和两根之积来做.能算出a+b和a*b的答案.请问多了log是怎么做出答案的?

由已知得：
log2^a+log2^b=2--------------1
log2^a*log2^b=-2-------------2

那么 由1得：
lga/lg2 + lgb/lg2=2

(lga+lgb)=2lg2
由2得：
(lga/lg2)*(lgb/lg2)=-2
lga *lgb=-2(lg2)^2
因此 loga^b+logb^a

= lga/lgb + lgb/lga
= [(lga)^2+(lgb)^2]/(lga *lgb)
= [(lga+lgb)^2-2lga*lgb]/(lga*lgb)
= (lga+lgb)^2/(lga*lgb) - 2
= (2lg2)^2/[-2(lg2)^2] - 2
= -2-2
=-4
应该很详细了,能看懂么