# 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.

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

• okay , Can you say final answer ? – S.H.W Jan 7 '17 at 9:24
• Sir $\lfloor \ln x \rfloor =-1$. – Rohan Jan 7 '17 at 9:34
• Yes. I edited!! – Roman83 Jan 7 '17 at 9:35

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.

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

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

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.