The question requires finding all real values of $x$ for which $$\lfloor \ln x\rfloor \gt \ln\lfloor x\rfloor $$ To start off, one could note that $$\lfloor \ln x \rfloor =\begin{cases} 0,& x\in[1,e) \\ 1,& x\in[e,e^2) \\ 2, &x\in [e^2,e^3) \\ 3,& x\in [e^3,e^4) \\ \vdots \end{cases}$$ and
$$\ln\lfloor x\rfloor =\begin{cases} 0, &x\in[1,2) \\ \ln 2, &x\in [2,3) \\ \ln 3,& x\in[3,4) \\ \ln 4,& x\in[4,5) \\ \vdots \end{cases}$$Although, from here it is not exactly clear to me how I can proceed. It appeas to me that there are infinitely many intervals of $x$ for which this inequality is true, but how can I find a generalized form of such an interval? E.g. something of the form $x\in \big(f(k), g(k)\big)$ for $k\in\mathbb N$ ?
Best Answer
You have a solution in every interval $[k, k+1), k \in \mathbb N$ which contains a power of $e.$
To prove this, consider breaking up $[k, k+1)$ as $[k, e^\alpha) \cup[e^\alpha, k+1).$
In the first interval, $\lfloor{\ln x}\rfloor = \alpha -1, $ while $\ln\lfloor{ x}\rfloor = \ln k$.
It is easy enough to see that the inequality doesn't hold in this region (Using the fact that $k > e^{\alpha - 1}$) For the second interval, $\lfloor{\ln x}\rfloor = \alpha, $ while $\ln\lfloor{ x}\rfloor = \ln k$.
Obviously, $\alpha > \ln k$ so the inequality holds.
Hence, your solution set is of the form: $[e^\alpha, \lceil{e^\alpha}\rceil) \quad\forall \alpha \in \mathbb N$