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what is result following forlmula?

$\sqrt{n} \log_2 n/n $

I saw in a book that achieve(get) to $ \sqrt{n} / \log_2 n$

i mean $\sqrt{n} \log_2 n/n = \sqrt{n} / \log_2 n $

but how to get this result?

in the book writen that $\sqrt{n} \log_2 n$ has less Exponential growth than $n \log_2 2 $ and after writen becuse in $ \sqrt{n} / \log_2 n$ is $ \sqrt{n}$ always gratter than $ \log_2 n$. (you can removing $\log_2 2$ beacuse that is a constant)

If you don't understand some part, then don't give a negative veto, please help me to improve it or delete .. please! (my English lan isn't very good and I need StackOverflow).

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    $\begingroup$ $\sqrt{n} \log_2(n)/n=\log_2(n)/\sqrt{n}$, your version dividing by the logarithm is just not right. $\endgroup$ – Ian Nov 5 '19 at 18:53
  • $\begingroup$ after getting the result the dividing there was written proof: $\sqrt{n} \log_2 n$ has fewer Function of Growth than $n$ $\endgroup$ – Michael Nov 5 '19 at 19:05
  • $\begingroup$ and was the proof because $ \sqrt{n} / \log_2 n$ is $ \sqrt{n} $ gratter than $\log_2 n$ $\endgroup$ – Michael Nov 5 '19 at 19:11
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No it is not true, we have that

$$ \frac{\sqrt{n}}n\log_2 n=\frac{\sqrt{n}}{\sqrt{n}}\frac{\sqrt{n}}n\log_2 n=\frac 1{\sqrt{n}}\log_2 n=\frac1{\frac{\sqrt{n}}{\log_2 n}}$$


Edit

What is true is that $n \log_2 2$ growth faster than $\sqrt{n} \log_2 n$ indeed

$$\frac{n \log_2 2}{\sqrt{n} \log_2 n}=\frac{\sqrt{n}}{\sqrt{n}}\frac{n \cdot 1}{\sqrt{n} \log_2 n}=\frac{\sqrt n}{\log_2 n}$$

and for $n$ large $\sqrt n > \log_2 n$.

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  • $\begingroup$ thanks a lot ... please again checking the question(the question is edited) $\endgroup$ – Michael Nov 5 '19 at 19:43
  • $\begingroup$ tanks you very much ... but I have one question: why $\sqrt{n} * n.1 = \sqrt{n}$ ? $\endgroup$ – Michael Nov 5 '19 at 20:01
  • $\begingroup$ The $n$ term cancels out with $\sqrt n \cdot \sqrt n=n$ at the denominator. $\endgroup$ – user Nov 5 '19 at 20:06
  • $\begingroup$ ok ... thank you so much. $\endgroup$ – Michael Nov 5 '19 at 20:07
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$$\sqrt{n} \log_2 n/n = \sqrt{n} / \log_2 n$$ is not correct.

The correct form should be $$\sqrt{n} \log_2 n/n = \frac { \log_2 n}{\sqrt{n} }$$

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  • $\begingroup$ after getting the result the dividing there was written proof: $\sqrt{n} \log_2 n$ has fewer Function of Growth than $n$ $\endgroup$ – Michael Nov 5 '19 at 19:05
  • $\begingroup$ Yes, that statement is correct because $\sqrt n$ dominates $\log _2 n$ $\endgroup$ – Mohammad Riazi-Kermani Nov 5 '19 at 19:33
  • $\begingroup$ thanks a lot ... please again checking the question(the question is edited) $\endgroup$ – Michael Nov 5 '19 at 19:43

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