Clarification with logarithms $n^{\log\log n} = (\log n)^{\log n}$.

Question

How come the following are all true?:

$n^{\log_2\log_2 n} = (\log_2 n)^{\log_2 n}$

$n^2 = 4^{\log_2 n}$

$n = 2^{\log_2 n}$

$2^{\sqrt {2\log_2 n}} = n^{\sqrt {2/\log_2 n}}$

$1 = n ^ {1/ \log_2 n}$

Answer 1

I think..The main identity in your problems..

$$a^{\log_bc}=c^{\log_ba}$$ S0 $1),2),3)$ are true...$4),5)$ is false try $n=1$ ..Moreover ,even if it was defined

$$n^{1/\log_2n}=n^{\log_n2}=2\neq1$$

EDIT:

The identity itself can be proved easily

$$\log_ac=\frac{\log_bc}{\log_ba}$$

$$\log_ac\cdot\log_ba=\log_bc\cdot\log_aa$$ $$\log_ac^{\log_ba}=\log_aa^{\log_bc}$$ $$c^{\log_ba}=a^{\log_bc}$$

Answer 2

Let $$x=n^{\log_2(\log_2 n)}$$ Taking $\log_2$ on both sides, $$\begin{array}{rcl} \log_2 x &=& \log_2\left[n^{\log_2(\log_2 n)}\right] \\ &=& \log_2(\log_2 n)\cdot\log_2 n \\ &=& \log_2\left[\left(\log_2 n\right)^{\log_2 n}\right] \end{array}$$ Therefore $$x=\left(\log_2 n\right)^{\log_2 n}$$ This answers your first question. Your other questions except the last one are basically arithmetic; for example $n=2^{\log_2n}$ is just saying that $\log_2n=\log_2n$. Your last statement however appears not to be true; I think you mean $$2=n^{1/\log_2n}$$