已知n属于N,求证:logn(n+1)大于logn+1(n+2)

Question

logn(n+1)代表以n为底数的对数logn+1(n+2)代表以n+1为底数的对数

Answer 1

要证log(n)(n+1)>log(n+1)(n+2),
即要证lg(n+1)/lgn>lg(n+2)/lg(n+1)(换底公式)
即要证lg(n+1)lg(n+1)>lgnlg(n+2)
而由基本不等式
lgnlg(n+2)<((lgn+lg(n+2))/2)^2=(0.5lgn(n+2))^2
=(0.5lg(n+1-1)(n+1+1))^2
<(0.5lg(n+1)^2)^2
=lg(n+1)lg(n+1) 即log n (n+1)>log n+1(n+2)(移项)