TITLE: Solve floor equation over real numbers: $\lfloor x \rfloor + \lfloor -x \rfloor = \lfloor \log x \rfloor$
QUESTION [5 upvotes]: Consider : $\lfloor x \rfloor + \lfloor -x \rfloor = \lfloor \log x \rfloor$. How we can solve it over real numbers?
My try : I tried to solve it in several intervals but didn't get any result.
Please Help!

REPLY [2 votes]: The range of $[x]+[-x]$ will be $\{-1,0\}$. If $x$ is integer then $0$, if not integer then $-1$.
$[\log x]$ should be $\{-1,0\}$ $\Rightarrow$ $\log x$ can take values $[-1,1)$ $\Rightarrow$ $x$ can take values from $[e^{-1},e)$.
Therefore the solution is $[e^{-1},1]\bigcup\{2\}$.