Solve floor equation over real numbers: $\lfloor x \rfloor + \lfloor -x \rfloor = \lfloor \log x \rfloor$

Question

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!

Answer 1

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

Case 2) $$\lfloor \ln x \rfloor=-1$$ $$-1\le \ln x<0$$ $$\frac 1e\le x<1 $$

Answer 2

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.

Answer 3

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.

Answer 4

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. $-1 \le \log x < 0$

Answer 5

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 $[e^{-1},1]\bigcup\{2\}$.