Let we have the following equation with the unknown $x$ $$\lfloor\ln(x+1)\rfloor - \lfloor\ln(x)\rfloor = 1$$ Where $\lfloor x\rfloor$ means the integer part of $x$

  • $\begingroup$ Use formula for difference of logs. $\endgroup$ – coffeemath Jan 1 '19 at 10:09
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    $\begingroup$ Why did you enclose in square parentheses [ ] each one of those expressions? Why not merely write $\;\ln(x+1) - \ln x\;$ ? Does this mean anything special? And after this, what properties of logarithms you know? $\endgroup$ – DonAntonio Jan 1 '19 at 10:09
  • $\begingroup$ It is [x] integer part of x, I think $\endgroup$ – Aqua Jan 1 '19 at 10:11
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    $\begingroup$ Nice. The OP hasn't even addressed the comment, but someone already decided that he can read minds and edited the question... $\endgroup$ – DonAntonio Jan 1 '19 at 11:22
  • $\begingroup$ but there is two answers already. $\endgroup$ – onepound Jan 1 '19 at 13:55

The solution set is a sequence of intervals. For each positive integer $N$, we can construct an interval containing $e^N$. Let

$$\epsilon = \ln\left(\frac{e^N}{e^N+1}\right).$$

Then the interval from $e^{N-\epsilon}$ to $e^N$ is a set of solutions.

You need

$$\ln(e^{N-\epsilon}+1)\geq N$$

and solving this inequality gives the value for $\epsilon.$


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