x^2=2^x
May 10th, 2016 at 3:24:36 AM
permalink
Solutions are 2, 4, and an approximate solution is -(0.7+.2/3) which has an error of approximately 0.0007%
What are the exact solutions?
May 10th, 2016 at 3:36:28 AM
permalink
2^x = x^2
(2^x)^2 = x
x = (2^x)^2
.
.
.
x = -(2 W_n(-(log(2))/2))/(log(2)), n element Z
(2^x)^2 = x
x = (2^x)^2
.
.
.
x = -(2 W_n(-(log(2))/2))/(log(2)), n element Z
May 10th, 2016 at 8:24:36 AM
permalink
Good one. I rearranged it to x^(2/x) = 2 but don't know what to do with it from there.
"My life is spent in one long effort to escape from the commonplace of existence. These little problems help me to do so." -- Sherlock Holmes
May 10th, 2016 at 3:11:05 PM
permalink
Quote: WizardGood one. I rearranged it to x^(2/x) = 2 but don't know what to do with it from there.
2 = x^(2/x)
Couldn't you change that to:
2 = x^([x^(2/x)]/x)
And then do a series thingie?
May 11th, 2016 at 8:24:08 AM
permalink
The power tower -{(sqrt(2)/2)^(sqrt(2)/2)^(sqrt(2)/2)^(sqrt(2)/2)^(sqrt(2)/2)^(sqrt(2)/2)^...} .
Which sort of number is this?
Which sort of number is this?
So much bullshit; so little time!
