Solve $lfloor ln x rfloor gt ln lfloor xrfloor$

Question

The question requires finding all real values of $x$ for which $$lfloor ln xrfloor gt lnlfloor xrfloor $$ To start off, one could note that $$lfloor ln x rfloor =begin{cases} 0,& xin[1,e) \ 1,& xin[e,e^2) \ 2, &xin [e^2,e^3) \ 3,& xin [e^3,e^4) \ vdots end{cases}$$ and
$$lnlfloor xrfloor =begin{cases} 0, &xin[1,2) \ ln 2, &xin [2,3) \ ln 3,& xin[3,4) \ ln 4,& xin[4,5) \ vdots end{cases}$$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 $xin big(f(k), g(k)big)$ for $kinmathbb N$ ?

Nikunj · 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 $lnlfloor{ x}rfloor = ln k$.
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 $lnlfloor{ x}rfloor = ln k$.
Obviously, $alpha > ln k$ so the inequality holds.
Hence, your solution set is of the form: $[e^alpha, lceil{e^alpha}rceil) quadforall alpha in mathbb N$

Solve $lfloor ln x rfloor gt ln lfloor xrfloor$

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