First off, why did this post get downvoted?
Answer alert
The answer is sqrt(2), because sqrt(2)^2^ = 2, and you build the infinite exponential sequence by taking x_n = x^x_(n-1)^.
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First off, why did this post get downvoted?
Answer alert
The answer is sqrt(2), because sqrt(2)^2^ = 2, and you build the infinite exponential sequence by taking x_n = x^x_(n-1)^.
yup there u go, btw can u please spoiler it so its harder for people to accidentally see it
Now try x^x^x^… = 4. And then tell me why it’s not a logical crisis that you get exactly the same answer when you solve x^x^x^… = 2 instead.
The reason is that x^x^... doesn't converge to anything greater than e.
Here's what happens with sqrt(2) as x with a 12 large power tower (WolframAlpha doesn't like power towers with roots):
Here's what happens with e^(1/e)^, the max value for x with a 500,000 large power tower:
(Note: I was unable to enter that value so I approximated it. The exact result should be e)
Here's what happens with the value above +0.00001 and a mere 1023 large power tower:
You can read the reasons here, I couldn't follow for very much but maybe you can get something out of it:
What is weird is that x^x^x^x^.... does have multiple results as for e.g. x=√2 it can yield 2, 4, and possibly more. Maybe it is due to the concept of infinity that we cannot really grasp.
spoiler
Allow y = x^x^x^...
Notice the exponent tower contains itself, such that:
y = x^y
Raise both sides to 1/y power:
y^(1/y) = x^(y/y)
y^(1/y) = x
For y = 2,
2^(1/2) = x
Aka:
x = √2