Deven Livingston

2022-04-26

What is intresting about

$\sqrt{{\mathrm{log}}_{x}\mathrm{exp}\sqrt{{\mathrm{log}}_{x}\mathrm{exp}\sqrt{{\mathrm{log}}_{x}\mathrm{exp}\cdots}}}={\mathrm{log}}_{x}e$

Why does

$\sqrt{{\mathrm{log}}_{x}\mathrm{exp}\sqrt{{\mathrm{log}}_{x}\mathrm{exp}\sqrt{{\mathrm{log}}_{x}\mathrm{exp}\cdots}}}={\mathrm{log}}_{x}e=\frac{1}{\mathrm{ln}x}$

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 ${\mathrm{log}}_{x}a)$ .

Why does

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

Giovanny Howe

Beginner2022-04-27Added 18 answers

We have an $n$ th root instead of a square root:

$\sqrt[n]{{\mathrm{log}}_{x}\mathrm{exp}\sqrt[n]{{\mathrm{log}}_{x}\mathrm{exp}\sqrt[n]{{\mathrm{log}}_{x}\mathrm{exp}\cdots}}}={\mathrm{log}}_{x}e=\frac{1}{\mathrm{ln}x}$

Then

$y=\sqrt[n]{{\mathrm{log}}_{x}\mathrm{exp}\left(y\right)}$

$y=\sqrt[n]{y{\mathrm{log}}_{x}e}$

${y}^{n}=y{\mathrm{log}}_{x}e$

${e}^{n-1}={\mathrm{log}}_{x}e$

Obviously, with $n=2,n-1=1$, meaning $y$ itself equals ${\mathrm{log}}_{x}e$.

This can be expanded upon though.

Tyler Velasquez

Beginner2022-04-28Added 19 answers

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