0
$\begingroup$

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 $x$

$\endgroup$
  • $\begingroup$ Use formula for difference of logs. $\endgroup$ – coffeemath Jan 1 '19 at 10:09
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
2
$\begingroup$

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 $\epsilon.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.