Question

一道高中数学题 高手进

已知log5(√6+1）+log2(√2-1）=a，则log5(√6-1）+log2(√2+1）=

Answer 1

设：log5(√6-1）+log2(√2+1）= b
a+b=log5(√6+1）+log2(√2-1）+log5(√6-1）+log2(√2+1）
=log5〔(√6+1）(√6-1）〕+log2〔(√2-1）(√2+1）〕
=log5（6-1）+log2（2-1）
=1+0=1
所以log5(√6-1）+log2(√2+1）= b=1-a

Answer 2

注意到 (-1+√6)(√6+1)=6-1=5 => -1+√6=5/(1+√6)
(1+√2)(-1+√2)=2-1=1 => 1+√2 =1/(-1+√2)
log5(-1+√6)+log2(1+√2)
=log5[5/(1+√6)] + log2[1/(-1+√2)]
=log5(5)-log5(1+√6)-log2(-1+√2)
=1-a