I have to check if $(\sqrt{2})^{\lg n} = 2^{\sqrt{2\lg n}}$. My idea was to take logs: $\lg\ (\sqrt{2})^{\lg n} =\lg(2^{\sqrt{2\lg n}})$. But how to simplify further? What should I do next? Please, explain in details if possible.
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The left-hand side is $2^{(\log n)/2}$. The right-hand side is $2^{\sqrt{2\log n}}$. The only time these are equal is when $(\log n) / 2 = \sqrt{2\log n}$, which is when $\log n = 8$. – mjqxxxx Sep 12 '14 at 21:17
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They're not equivalent. For example, take $n = 4$. Then assuming that $\lg x \equiv \log_2 x$, notice that: $$ (\sqrt 2)^{\lg 4} = (\sqrt 2)^2 = 2 \neq 4 = 2^{2} = 2^{\sqrt{4}} = 2^{\sqrt{2(2)}} = 2^{\sqrt{2 \lg 4}} $$
Adriano
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$$ \sqrt{2}\,^{\log{n}} = (2^{\tfrac{1}{2}})^{\log{n}} = 2^{{\tfrac{1}{2}}\log{n}} = 2\,^{\log{\sqrt{n}}}= (2^{\log{n}})^{\tfrac{1}{2}} = \sqrt{2^{\log{n}}} \\ =n^{\tfrac{1}{2}log{\,2}} = n^{\log{\sqrt{2}}} = \sqrt{n}^{\,\log{2}}= \sqrt{n^{\log{2}}} $$
All of these can be derived from the rules $$ \log{a^b} = b\log{a} \\a^{bc} = (a^{b})^c $$ but they are not equivalent to your RHS expression.
MartinG
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