x^2=2^x
May 10th, 2016 at 3:24:36 AM
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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
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2^x = x^2
(2^x)^2 = x
x = (2^x)^2
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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
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Good one. I rearranged it to x^(2/x) = 2 but don't know what to do with it from there.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
May 10th, 2016 at 3:11:05 PM
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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
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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!
