Why does $$\sqrt{\log_x{\exp{\sqrt{\log_x{\exp{\sqrt{\log_x{\exp{\cdots}}}}}}}}}=\log_x{e}=\frac{1}{\ln{x}}$$ There only seems to be a relation when using square roots, but not for cubed roots or anything else. Why does this equation work and why does it only work for square roots?

(The $e$ is not significant, by the way. You could give the exponential function a different base, $a$, and say the equation equals $log_x{a}$).

  • $\begingroup$ Call the expression $A$; if the expression exists, we would have $A=\sqrt{\log_x\exp A}$, right? $\endgroup$ Aug 11, 2015 at 1:47
  • $\begingroup$ holly hell, where did this come from? +1 $\endgroup$
    – jimjim
    Aug 11, 2015 at 1:50
  • $\begingroup$ If the starting value of the recursion is $0$ then the limit is $0$ since $\log_x \exp(0) = 0$. $\endgroup$
    – Winther
    Aug 11, 2015 at 2:10
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    $\begingroup$ The answers are the to the content, your title and content are different. Title is asking what , content is asking why? I started a more specific question regarding what is interesting math.stackexchange.com/questions/1392673/… $\endgroup$
    – jimjim
    Aug 11, 2015 at 2:23
  • $\begingroup$ @Winther I think $\log_0x=0$ would make a reasonable definition. After all, $\lim_{b\to0^+}\log_bx=0$. Also, it's what Wolfram Alpha does. $\endgroup$ Aug 11, 2015 at 4:44

2 Answers 2


Elaborating on what Jack said, assume we have an $n$th root instead of a square root:

$$y = \sqrt[n]{\log_x{\exp{\sqrt[n]{\log_x{\exp{\sqrt[n]{\log_x{\exp{\cdots}}}}}}}}}$$


$$y = \sqrt[n]{\log_x{\exp\left(y\right)}}$$

$$y = \sqrt[n]{y\log_x{e}}$$

$$y^n = y\log_x{e}$$

$$y^{n-1} = \log_xe$$

Obviously, with $n = 2$, $n-1 = 1$, meaning $y$ itself equals $\log_xe$.

This can be expanded upon though.


$$y=\sqrt{\log_x{\exp{\sqrt{\log_x{\exp{\sqrt{\log_x{\exp{\cdots}}}}}}}}}\implies y=\sqrt{\log_x\exp(y)}=\sqrt{y\log_xe}\\ \therefore y=\log_xe$$

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    $\begingroup$ Or $y=0$ which happens if the starting value is $0$. $\endgroup$
    – Winther
    Aug 11, 2015 at 2:12
  • $\begingroup$ @Winter You're right but my head hurts when I try to think about the "starting value" :D $\endgroup$ Aug 11, 2015 at 2:14
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    $\begingroup$ Hehe. Maybe easier to think of it as a recursion $y_{n+1} = \sqrt{\log_x\exp(y_n)}$ then $y = \lim_{n\to\infty} y_n$. $\endgroup$
    – Winther
    Aug 11, 2015 at 2:17

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