若log7 (2√2+1)+log2 (√2-1)=a,把log7 (2√2-1)+log2 (√2+1)表示为a的代数式

问题描述:

若log7 (2√2+1)+log2 (√2-1)=a,把log7 (2√2-1)+log2 (√2+1)表示为a的代数式

两式相加
log7 (2√2+1)+log2 (√2-1)+log7 (2√2-1)+log2 (√2+1)
=log7 [(2√2+1)(2√2-1)] +log2 [(√2-1)(√2+1)]
=log7 7 + log2 1
=1+0=1
又log7 (2√2+1)+log2 (√2-1)=a
所以
log7 (2√2-1)+log2 (√2+1)=1-a