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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

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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

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  • $\begingroup$ 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

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