4
$\begingroup$

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 $log_x{a}$).

$\endgroup$
7
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

10
$\begingroup$

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

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

Then

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

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

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

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


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

This can be expanded upon though.

$\endgroup$
8
$\begingroup$

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

$\endgroup$
3
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.