# 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

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)

• you likely want $\ln x$ in the last inequality? – gt6989b Dec 2 '14 at 18:39
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 $$x=3$$.