# what is result $\sqrt{n} \log_2 n/n$?

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).

• $\sqrt{n} \log_2(n)/n=\log_2(n)/\sqrt{n}$, your version dividing by the logarithm is just not right. – Ian Nov 5 '19 at 18:53
• after getting the result the dividing there was written proof: $\sqrt{n} \log_2 n$ has fewer Function of Growth than $n$ – Michael Nov 5 '19 at 19:05
• and was the proof because $\sqrt{n} / \log_2 n$ is $\sqrt{n}$ gratter than $\log_2 n$ – Michael Nov 5 '19 at 19:11

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 $$\sqrt n > \log_2 n$$.

• thanks a lot ... please again checking the question(the question is edited) – Michael Nov 5 '19 at 19:43
• tanks you very much ... but I have one question: why $\sqrt{n} * n.1 = \sqrt{n}$ ? – Michael Nov 5 '19 at 20:01
• The $n$ term cancels out with $\sqrt n \cdot \sqrt n=n$ at the denominator. – user Nov 5 '19 at 20:06
• ok ... thank you so much. – Michael Nov 5 '19 at 20:07

$$\sqrt{n} \log_2 n/n = \sqrt{n} / \log_2 n$$ is not correct.

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

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