Calculation of all positive integer $x$ for which $\displaystyle \lfloor \log_{2}(x) \rfloor = \lfloor \log_{3}(x) \rfloor \;,$
where $\lfloor x \rfloor $ represent floor function of $x$.
$\bf{My\; Try::}$ I have used the fact that $\lfloor x\rfloor = \lfloor y \rfloor\;,$ is possible when $x,y\in \left[k\;,k+1\right)\;,$
where $k\in \mathbb{Z}$ and $\left|x-y\right|<1.$
So $\displaystyle \left|\log_{2}(x)-\log_{3}(x)\right|<1\Rightarrow -1<\log_{2}(x)-\log_{3}(x)<1$
Now how can I calculate it, Help me
thanks