Why does $2^{\log\left(n\right)\left(\log\left(\log\left(n\right)\right)\right)} = n^{\log\left(\log\left(n\right)\right)}$?

Question

There are a few ways to simplify the expression: $\left(\log\left(n\right)\right)^{\log\left(n\right)}$

Edit: The Log is BASE 2.

One way, would be using the rule $a^{\log\left(b\right)}=b^{\log\left(a\right)}$

So, we would have the equation $\left(\log\left(n\right)\right)^{\log\left(n\right)}=\:n^{\log\left(\log\left(n\right)\right)}\:$

But there is another way that I didn't really understand.

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We will append a log to this expression $\left(\log\left(n\right)\right)^{\log\left(n\right)}$, so it will be $\log\left(\log\left(n\right)^{\log\left(n\right)}\right)$

according to the rule $\log(a)^b = b\log(a)$

it is going to be $\log(n)(\log(\log(n)))$

We are going to do the opposite of the $\log$ operation, and make it an exponent for the base 2:

$2^{\log\left(n\right)\left(\log\left(\log\left(n\right)\right)\right)}$

The step I didn't understand is.. why does $2^{\log\left(n\right)\left(\log\left(\log\left(n\right)\right)\right)} = n^{\log\left(\log\left(n\right)\right)}$ ?

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Thanks for reading.

Answer 1

So, you reach the point where you got $$ (\log n)^{\log n} = 2^{\log(n)\cdot \log\log n} $$ (again, $\log$ here is base $2$).

Now, all you need to remember is that (1) $a^{bc} = (a^b)^c$ and (2) $2^{\log a} = a$ (for $a>0$), so $$ (\log n)^{\log n} = 2^{\log(n)\cdot \log\log n} = \left(2^{\log(n)}\right)^{\log\log n} = \boxed{n^{\log\log n}} $$