Calculation of all positive integer $x$ for which $\lfloor \log_{2}(x) \rfloor = \lfloor \log_{3}(x) \rfloor \;,$ 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
 A: Log base 2 of x = lnx/ln2 and base 3 of x = lnx/ln3
Multiply the equation by ln2:
-ln2

Multiply by ln3:
-ln2ln3

Divide by (ln3-ln2)
-(ln2ln3)/(ln3-ln2) < x < (ln2ln3)/(ln3-ln2)
Sorry for the bad format
A: Let $n$ be an integer such that $\lfloor \log_2 x \rfloor = n = \lfloor \log_3 x \rfloor$. If $n=0$, then $x=1$ is a solution. Suppose that $n\ge 1$. By the definiton of the floor function, we obtain
$$
\begin{cases}
2^n \le x < 2^{n+1}\\
3^n \le x < 3^{n+1}
\end{cases}
$$
It is obvious that $2^n < 2^{n+1} < 3^{n+1}$ and $2^n < 3^n < 3^{n+1}$, but it is uncertain to compare $2^{n+1}$ and $3^n$. If $2^{n+1}\le 3^n$, then there is no solution. Suppose that $2^{n+1}>3^n$, then
\begin{align}
n+1 &> n\log_2 3\\
1+\frac{1}{n} &> \log_2 3\\
\therefore n&< \frac{1}{\log_2 3 -1} =\frac{1}{\log_2 \frac{3}{2}} < \frac{1}{\log_2 \sqrt{2}} =2
\end{align}
Thus $n$ must be $1$, and we obtain $x=3$.
Conclusion: All integer solutions of $\lfloor \log_2 x \rfloor = \lfloor \log_3 x \rfloor$ are $x=1$ and $x=3$.
