已知a大于2求证loga(a-1)

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已知a大于2求证loga(a-1)

已知a大于2求证loga(a-1)
已知a大于2求证loga(a-1)

已知a大于2求证loga(a-1)
a>2 a-1>1 a+1>1
loga(a-1)>0 loga(a+1)>0
先证loga(a-1)

由mn<=[(m+n)/2]^2得
log(a-1)*log(a+1)<={[log(a-1)+log(a+1)]/2}^2
={log
[(a-1)(a+1)]/2}^2
={[log
(a^2-1)]/2}^2
<{[log
(a^2)]/2}^2=1,
a>2,∴0(a+1),上式变为
log
(a-1)<1/log(a+1),换底得
log
(a-1)a.