已知log5(根号6+1)+log2(根号2-1)=a,则log5(根号6-1)+log2(根号2+1)=
问题描述：
已知log5(根号6+1)+log2(根号2-1)=a,则log5(根号6-1)+log2(根号2+1)=
答
已知log‹5›(√6+1)+log‹2›(√2-1)=a，则log‹5›(√6-1)+log‹2›(√2+1)=
log‹5›(√6+1)+log‹2›(√2-1)=log‹5›[5/(√6-1)]+log‹2›[1/(√2+1)]
=1-log‹5›(√6-1)-log‹2›(√2+1)=a
故 log‹5›(√6-1)+log‹2›(√2+1)=1-a
答
log5(√6+1)+log2(√2-1)+log5(√6-1)+log2(√2+1)
=log5[(√6+1)(√6-1)]+log2[(√2-1)(√2+1)]
=log5(5)+log2(1)
=1
log5(√6-1)+log2(√2+1)=1-a