Is it possible to simplify $\log_2(\log_2(x))$ to a single log of some base, and possibly some power or multiplier for x? Or some other way? Or anything that doesn't involve a double log?

  • $\begingroup$ If so: no. Changing the base of the logarithm will make a difference by a constant factor; while $\log_2\log_2 x$ is exponentially smaller than $\log_2 x$. $\endgroup$ – Clement C. Mar 6 '17 at 19:24
  • $\begingroup$ Thanks. But is there another way - like I also ask for? E.g raising the log2 to some power? $\endgroup$ – Matceporial Mar 6 '17 at 19:24
  • $\begingroup$ Note sure exactly what you are asking for, but can you use $$2^{\log_2(x)} = x?$$ $\endgroup$ – gt6989b Mar 6 '17 at 19:27
  • $\begingroup$ Nope, this is about as simple as it gets. $\endgroup$ – Michael Burr Mar 6 '17 at 19:28
  • $\begingroup$ Try any way that doesn't involve double logs $\endgroup$ – Matceporial Mar 6 '17 at 19:32

Suppose that $y=\log_2(\log_2x)$.

Then $2^y=\log_2x$ and $2^{(2^y)}=x$.

Essentially what OP is asking is whether there exists a base $b$ such that $b^y=x$. Let us suppose there were.

Let $b^y=x=2^{(2^y)}$. Then $y=\log_b2^{(2^y)}=2^y\log_b2$.

Therefore, $\log_b2=y\cdot 2^{-y}$

But $\log_b2$ is a constant and $y\cdot 2^{-y}$ is not a constant.

So there can be no base $b$ such that $b^y=2^{(2^y)}$ and therefore no way to simplify $\log_2(\log_2y)$ to some $\log_by$.


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