Arithmetically showing that \(\displaystyle{\frac{{{\log{{\left({x}+{1}\right)}}}}}{{{\log{{\left({x}\right)}}}}}}{ < }{\frac{{{x}+{1}}}{{{x}}}}\) Is there

Babolkybya 2022-03-15 Answered
Arithmetically showing that log(x+1)log(x)<x+1x
Is there a possibility that this can be shown arithmetically? By arithmetically, I mean not looking at the graph.
log(x+1)log(x)<x+1x
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Answers (2)

alifutlessect692
Answered 2022-03-16 Author has 9 answers
The derivative of xlnxx is 1lnxx2 and this is negative if x>e . Thus is 1<x<e1, we have ln(x+1)x+1>lnxx and after dividing by the positive number lnx and multiplying with x+1, this yields ln(x+1)lnx>x+1x
However, for x>e (and even some smaller x) we obtain ln(x+1)lnx<x+1x
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Loyau97
Answered 2022-03-17 Author has 2 answers
Your statement is false in general, as already pointed out. However, the following inequality holds
log(x+1)=log(x)+log(1+1x)log(x)+1x.
So if x>e then
log(x+1)log(x)1+1xlog(x)<1+1x.
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