Solve floor equation over real numbers: $\lfloor x \rfloor + \lfloor -x \rfloor = \lfloor \log x \rfloor$ 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! 
 A: We know that if $x \in \mathbb Z$
$$\lfloor x \rfloor + \lfloor -x \rfloor =0$$
And if $x \not\in \mathbb Z$
$$\lfloor x \rfloor + \lfloor -x \rfloor =-1$$
So we can easily see the solutions are $x \in \mathbb Z \in [1,e) $ and $x \not\in \mathbb Z \in [\frac {1}{e},1) $. Hope it helps. 
A: Hint:
note that:
$
\lfloor x\rfloor+\lfloor -x\rfloor=0
$
if $x \in \mathbb{Z}$, and 
$
\lfloor x\rfloor+\lfloor -x\rfloor=-1
$
if $x$ is not an integer.
A: For each integer value the LHS equals $0$, for each non-integer value it's $-1$.
So find integer values s.t. $0 \le \log x < 1$ and find all non-integers s.t. $-1 \le \log x < 0$
A: 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 $[e^{-1},1]\bigcup\{2\}$.
A: Hint:
If $x \in \mathbb Z$
$$\lfloor x \rfloor + \lfloor -x \rfloor =0$$
If $x \not\in \mathbb Z$
$$\lfloor x \rfloor + \lfloor -x \rfloor =-1$$
Case 1) $$\lfloor \ln x \rfloor=0$$
$$0\le \ln x<1$$
$$1\le x<e$$
$$x \in \{1,2\}$$
Case 2) $$\lfloor \ln x \rfloor=-1$$
$$-1\le \ln x<0$$
$$\frac 1e\le x<1 $$
