Real Floor Solutions

Given that

$\begin{aligned} \ln^2 x - \lfloor \ln x \rfloor & = 2 \\ \implies \ln^2 x & = \lfloor \ln x \rfloor + 2 & \small \blue{\text{Let }x = e^u} \\ \ln^2 \left(e^u \right) & = \lfloor \ln \left(e^u \right) \rfloor + 2 \\ u^2 & = \lfloor u \rfloor + 2 \end{aligned}$

Since the left-hand side, $u^2 \ge 0$ , the right-hand side, $\lfloor u \rfloor + 2 \ge 0 \implies u \ge -2$ . Since $\lfloor u \rfloor + 2$ is an integer, $u^2$ must also be an integer and $u$ is of the form $u = \pm \sqrt n$ , where $n$ is a positive integer. Then we have $n = \lfloor \pm \sqrt n \rfloor + 2$ . We note that $n = 1 \ (u = - 1)$ , $n = 3$ , and $n = 4$ are the $\boxed 3$ solutions. For $n \ge 5$ , $n > \lfloor \sqrt n \rfloor + 2$ and there is no solution.

To simplify things, let $y=\ln x$ . Then we have $y^2-\lfloor y\rfloor=2$ Since $y\geq\lfloor y\rfloor$ , replacing $\lfloor y\rfloor$ with $y$ gives the inequality $y^2-y\leq2$ $\implies (y-2)(y+1)\leq0$ $\implies -1\leq y\leq2$ So, we only need to check the following cases:

• When $\lfloor y\rfloor=2$ , we have $y^2=4$ , and the only solution satisfying $\lfloor y\rfloor=2$ is $y=2$
• When $\lfloor y\rfloor=1$ , we have $y^2=3$ , and the only solution satisfying $\lfloor y\rfloor=1$ is $y=\sqrt 3$
• When $\lfloor y\rfloor=0$ , we have $y^2=2$ , and no solutions to this satisfy $\lfloor y\rfloor=0$
• When $\lfloor y\rfloor=-1$ , we have $y^2=1$ , and the only solution satisfying $\lfloor y\rfloor=-1$ is $y=-1$

So, there are 3 solutions: $\ln x=2$ , $\ln x=\sqrt 3$ , and $\ln x=-1$ .