Question

已知a大于2求证loga(a-1)<log(a+1)a

loga(a-1)是以a为底(a-1)的对数，另一个同上述，用比较法证明

Answer 1

a>2 a-1>1 a+1>1
loga(a-1)>0 loga(a+1)>0

先证loga(a-1)<loga(a+1)

由均值不等式，得
loga(a-1)loga(a+1)<[loga(a-1)+loga(a+1)]^2/4
=loga[(a-1)(a+1)]^2/4
=loga(a^4)/4
=4/4
=1
因此loga(a-1)loga(a+1)<1
即
loga(a-1)<1/loga(a+1)
loga(a-1)<log(a+1)a

Answer 2

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