How to show that $2^{\sqrt{n}} = n^{\frac{\sqrt{n}}{\log_2 n}}$?

Question

How to show that $2^{\sqrt{n}} = n^{\tfrac{\sqrt{n}}{\log_2 n}}$?

Well, I've no idea how to begin solving it, and I'm struggling every day with these questions about logairthms in my course, can anyone give me a hint how to handle these kind of questions?

Answer 1

Answer to the original question

Prove that $(\log\log n)^{\log n} = n^{\tfrac{\sqrt{n}}{\log n}}$

Take the logarithm of the right-hand side: $$ \log\bigl(n^{\tfrac{\sqrt{n}}{\log n}}\bigr)= \frac{\sqrt{n}}{\log n}\log n=\sqrt{n} $$ so the right-hand side is $e^{\sqrt{n}}$ which is much greater, for large $n$, than the left-hand side.

Answer to the modified question

Prove that $2^{\sqrt{n}} = n^{\tfrac{\sqrt{n}}{\log n}}$

If $\log$ denotes the logarithm in base $a$, the same argument as before shows $$ n^{\tfrac{\sqrt{n}}{\log n}}=a^{\sqrt{n}} $$ If your base is $2$, then you get $2^{\sqrt{n}}$. However, writing $\log n$ to denote the logarithm in base $2$ is, at least, not standard.

Answer 2

Not equal.

Take $n=e$.

Then

$$(\log \log e)^{\log e}=0^1=0$$

and

$$e^{\frac{\sqrt e}{\log e}}=e^{\sqrt e}\ne 0.$$

Answer 3

Since $a^b=c^{b\log_c(a)}$ then it follows that $$n^{\frac{\sqrt{n}}{\log n}}=2^{\frac{\sqrt{n}}{\log_2 n}\cdot\log_2 n}=2^{\sqrt{n}}.$$