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Can anybody help me please to prove this equation?

$2^ \sqrt{2\lg n}$ = $n^ \sqrt{\left(\frac{2}{\lg n}\right)}$

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I suspect this is supposed to be for n>1? –  Mark Bennet Sep 12 '12 at 11:16

4 Answers 4

up vote 2 down vote accepted

Take $Log_2()$ of both sides:
$\sqrt2 \times \sqrt{\log_2(n)} \times \log_2(2) = \frac{\sqrt2}{\sqrt{\log_2(n)}}\times \log_2(n) $
$\Leftarrow$$Note:\log_2(2)=1 $
$\Leftarrow $Also note that $\frac{2}{\log_2n} >0 \Leftarrow {\log_2n}>0 \Leftarrow n>1$

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I was editing this . Concise and clear. (+1) –  Chris's sis Sep 12 '12 at 11:18
I guess the last series of implications is used backwards? –  Did Sep 13 '12 at 7:58
@did $n>1 \implies ... $ and the rest. Is it what you mean? –  Zeta.Investigator Sep 13 '12 at 8:01
This is what I mean and what you need. –  Did Sep 13 '12 at 8:09
@did I think this post is supposed to be a HINT for OP not a formal proof. –  Zeta.Investigator Sep 13 '12 at 8:12

Hint: Take the $\lg$ of both sides remembering that $\lg(x^y)=y\lg(x)$ and that $\lg(2)=1$.

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$2^\sqrt{2\log(n)}=(n^{\log(2) \over \log(n)})^{\sqrt{2\log(n)}}=n^{{\log(2) \over \log(n)}\sqrt{2\log(n)}}=n^{\sqrt{{\log(2)^2 \over \log(n)^2}2\log(n)}}=n^{\sqrt{{2\log(2)^2 \over \log(n)}}}$

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By definition, $a^b=\exp(b\ln a)$ for every $a\gt0$ hence the LHS is $\exp(\sqrt{2\lg n}\ln2)$ and the RHS is $\exp(\ln n\sqrt{2/\lg n})$. To conclude, note that $\ln n=\ln2\cdot\lg n$.

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