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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?

Matceporial
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  • 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 $\log_2 x$. – Clement C. Mar 06 '17 at 19:24
  • Thanks. But is there another way - like I also ask for? E.g raising the log2 to some power? – Matceporial Mar 06 '17 at 19:24
  • Note sure exactly what you are asking for, but can you use $$2^{\log_2(x)} = x?$$ – gt6989b Mar 06 '17 at 19:27
  • Nope, this is about as simple as it gets. – Michael Burr Mar 06 '17 at 19:28
  • Try any way that doesn't involve double logs – Matceporial Mar 06 '17 at 19:32
  • If you have a term of the form $log(x)log(log(x))$ then it can be "simplified" in terms of the Lambert W function. You can also simplify $log(log(x)$. For example $log(x)log(log(x)) = y$ is equivalent to $log(log(x)) = W(y)$ and to $log(x) = y/W(y)$ – Χpẘ Mar 06 '17 at 21:07

1 Answers1

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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 $\log_by$.

John Wayland Bales
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