What is intresting about \sqrt{\log_x\exp\sqrt{\log_x\exp\sqrt{\log_x\exp\cdots}}}=\log_xe

Deven Livingston

Deven Livingston

Answered question

2022-04-26

What is intresting about
logxexplogxexplogxexp=logxe
Why does
logxexplogxexplogxexp=logxe=1lnx
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 logxa).

Answer & Explanation

Giovanny Howe

Giovanny Howe

Beginner2022-04-27Added 18 answers

We have an n th root instead of a square root:
logxexplogxexplogxexpnnn=logxe=1lnx
Then
y=logxexp(y)n
y=ylogxen
yn=ylogxe
en1=logxe
Obviously, with n=2,n1=1, meaning y itself equals logxe.
This can be expanded upon though.

Tyler Velasquez

Tyler Velasquez

Beginner2022-04-28Added 19 answers

logxexplogxexplogxexp=logxey=logxexp(y)
=ylogxe
y=logxe

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?