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In a paper I was reading, this inequality: $$\log_{10}(1+10^{-n})<10^{-n}$$ came up with no explanation for why it's true. Does anyone have a proof for why this holds? Is there some basic logarithm property I'm missing?

Michael Hardy
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user2560855
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3 Answers3

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Yes, because:

  • $\ln(1+x)\leqslant x$ for every $x\gt-1$
  • $\log_{10}(t)=\ln(t)/\ln(10)$ by definition
  • $\ln(10)\gt1$ since $10\gt\mathrm e$

Hence, $\log_{10}(1+10^{-n})=\ln(1+10^{-n})/\ln(10)\lt10^{-n}/\ln(10)\lt10^{-n}$.

Did
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  • Would you really say that $\log_{10} t = \dfrac{\log_e t}{\log_e 10}$ is a DEFINITION? ${}\qquad{}$ – Michael Hardy Nov 11 '14 at 23:39
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    @MichaelHardy Yes. Wouldn't you? – Did Nov 11 '14 at 23:41
  • No. Mathematical logic hasn't caught up with practice here. "Rigor" may be characterized as either writing proof amenable to having their soundness checked algorithmically, or convincing oneself in each instance that one knows how to write such a proof. I think a characterization like that is treated as a definition only as a means to fit everything into that kind of concept of rigor. ${}\qquad{}$ – Michael Hardy Nov 12 '14 at 01:17
  • @MichaelHardy Sorry but you lost me. – Did Nov 14 '14 at 20:06
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For every number $x>-1$ $$ \log_e(1+x)\le x. $$

Therefore $$ \log_{10}(1+\varepsilon) = \frac{\log_e(1+\varepsilon)}{\log_e 10} \le \frac{\varepsilon}{\log_e 10} $$

And $\log_e 10>2$.

Michael Hardy
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If $f(x)=1+x-10^x$, note that if $x>0$, then $f'(x)<0$, then $f$ is decreasing, how $f(0)=0$, then $f(x)<0$: $$1+x-10^x<0\implies 1+x<10^x\implies \log_{10}(1+x)<x$$

AsdrubalBeltran
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