高二关于不等式和对数的题
n>1,n为整数,
证明:以n为底,(n+1)的对数大于以(n+1)为底,(n+2)的对数
参考答案:log(n+1)(n+2)/logn(n+1)
=[lgn*lg(n+2)]/[lg(n+1)]^2
lgn*lg(n+2)
<=[lgn+lg(n+2)/2]^2
=[lgn(n+2)/2]^2
又n(n+2)=n^2+2n<n^2+2n+1=(n+1)^2
<[lg(n+1)^2 / 2]^2
=[lg(n+1)]^2
即lgn*lg(n+2)<[lg(n+1)]^2
log(n+1)(n+2)/logn(n+1)
=[lgn*lg(n+2)]/[lg(n+1)]^2<1
结论得证