Checking a possible logarithm identity: $(\sqrt{2})^{\lg n} \stackrel{?}{=} 2^{\sqrt{2\lg n}}$

Question

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.

Answer 1

$$ \sqrt{2}\,^{\log{n}} = (2^{\tfrac{1}{2}})^{\log{n}} = 2^{{\tfrac{1}{2}}\log{n}} = 2\,^{\log{\sqrt{n}}}= (2^{\log{n}})^{\tfrac{1}{2}} = \sqrt{2^{\log{n}}} \\ =n^{\tfrac{1}{2}log{\,2}} = n^{\log{\sqrt{2}}} = \sqrt{n}^{\,\log{2}}= \sqrt{n^{\log{2}}} $$

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.

Answer 2

They're not equivalent. For example, take $n = 4$. Then assuming that $\lg x \equiv \log_2 x$, notice that: $$ (\sqrt 2)^{\lg 4} = (\sqrt 2)^2 = 2 \neq 4 = 2^{2} = 2^{\sqrt{4}} = 2^{\sqrt{2(2)}} = 2^{\sqrt{2 \lg 4}} $$