Smartass.
There is no such number.
There are only five different numbers that always produce the same result after powering with them for any number of times.
1, 0, -1, inf, -inf
The results of all of them are different than 2.
My answer is: "it's a trap"
To answer this problem you need to grasp what infinity means
"to grasp what infinity means", you must not be a finite being, but a god.
Here is my counter offer.
Tell me your number!
If it is correct, I will give you 1 SBD. If it is wrong, you will admit that I won your contest.
To make it easier to you, I will only test 3 versions.
x^x^x^x^x = 2
x^x^x^x^x^x^x = 2
x^x^x^x^x^x^x^x^x = 2
If all three equations are true for your number, I will give you 1 SBD.
Any real number is a valid entry. It doesn't matter if it is a floating point or an integer. As long as it is a number, I will accept it.
Why am I even talking to you?
You don't even have the 5$ you promised as reward.
Im buyin it for the person who gives me the correct answer
But if you want Iwill give you the answer
Yep. Give it.
I am willing to lose 1 SBD if you are right.
The answer is root 2
As I expected - you are wrong.
Before trying to be smart, you should understand what equality means.
(It is a lot simpler concept than infinity.)
Your formula will never, ever be equal to 2. It will always be slightly less, even if you apply it infinite times.
Mathematicians assume equality, because it is almost the same, but that doesn't mean it is really true.
I repeat myself - there is no number other than the 5 I listed, that always produces a constant after powering with it.
if you use proof it technically equals root 2. Doing root 2 ^ of root 2 it will equal 1.9 recurring which technically equals 2.
x = 1.9recurring
10x = 19.9(rec)
10x - x = 9x = 18
x therefore = 2