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



Log base 2 of x = lnx/ln2 and base 3 of x = lnx/ln3

Multiply the equation by ln2:


Multiply by ln3:


Divide by (ln3-ln2)

-(ln2ln3)/(ln3-ln2) < x < (ln2ln3)/(ln3-ln2)

Sorry for the bad format

  • $\begingroup$ you likely want $\ln x$ in the last inequality? $\endgroup$ – gt6989b Dec 2 '14 at 18:39
  • $\begingroup$ You can/should right-click a mathematical expression to see its LaTex format: [Show math as] --> [Tex commands]. An example from this question is: \displaystyle \lfloor \log_{2}(x) \rfloor = \lfloor \log_{3}(x) \rfloor \; $\endgroup$ – The Chaz 2.0 Dec 2 '14 at 18:39
  • $\begingroup$ @TheChaz2.0 I'm using an iPad nice tip though $\endgroup$ – Kurtbusch Dec 2 '14 at 18:45
  • $\begingroup$ Most of my questions/answers on here were from an iPad as well, so I understand the struggle! A workaround is to click "edit" on the question, then copy+paste relevant code. A litte "\log_{2}" would go a long way in your answer (wrapped in dollar signs, of course). $\endgroup$ – The Chaz 2.0 Dec 2 '14 at 20:46

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 $x=3$.


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