$\log_{10}(1+10^{-n})<10^{-n}$?

Question

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?

Answer 1

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

Answer 2

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 $\log_e 10>2$.

Answer 3

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 $f(x)<0$: $$1+x-10^x<0\implies 1+x<10^x\implies \log_{10}(1+x)<x$$