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x^(1/x) = a

Name: Anonymous 2006-01-06 21:55

x^(1/x) = a

find x in terms of a, K go!

Name: Anonymous 2006-01-07 1:14

ln(x^(1/x)) = ln(a)

ln(x)/x = ln(a)

ln(x) = x*ln(a)

d[ln(x)]/dx = d[x*ln(a)]/dx

1/x = ln(a)

x = 1/ln(a)

i win amirite. OGC

Name: Anonymous 2006-01-07 2:31

that transcendental equation has no analytic solution. nice try though!

Name: Anonymous 2006-01-07 7:22

FUCK OFF AND DO YOUR OWN HOMEWORK NOOB

Name: Anonymous 2006-01-07 10:29 (sage)

x^(1/x)=a
(x^(1/x))^x=a^x
x=a^x
x=a^(a^x)
x=a^(a^(a^x))
x=a^(a^(a^(a^x)))
infinite loop, OH SHI-

Name: Anonymous 2006-01-07 12:39

>>2
Congrats and wow, I didn't think any of you retards would get it. You are all monotonous drones who don't truly understand the nature of the equation you utilise. Nice thinking.

Your reward is 1 free 4chan. Enjoy!

Name: Anonymous 2006-01-07 12:56

actually you got it wrong

Name: Anonymous 2006-01-07 12:56

bullshit! you can't just derive like that!

Name: Anonymous 2006-01-07 13:24 (sage)

Yes you can, just fill it in:
(1/ln(a))^(1/(1/ln(a)))=a
(1/ln(a))^(ln(a))=a
ln(a)^(-ln(a))=a
Which is obviously true.

Name: Anonymous 2006-01-07 14:51

taking a=1 instantly disproves the x=1/log(a) hypothesis.

Name: Anonymous 2006-01-07 14:58

>>9
you're making me horny

Name: Anonymous 2006-01-07 15:19

>>11
Cool. Wanna get together?

Name: Anonymous 2006-01-07 15:33

>>4 wow, that's actually the correct answer... dumb fags

Name: Anonymous 2006-01-07 16:09

>>6
I got it wrong, retard. a is a variable.

Name: Anonymous 2006-01-07 20:20

fucking americans dont know math for shit no wonder your the fattest AND stupidest country in the world

Name: Anonymous 2006-01-07 23:05

SOMEONE GET A PROFESSOR

Name: Anonymous 2006-01-08 2:56

SOMEONE GET THIS THREAD A TROLL SWEEP

WOOP WOOP WOOP WOOP WOOP!

Name: Anonymous 2006-01-08 16:13

pls answer pls

Name: Anonymous 2006-01-09 3:12

there is no answer

Name: Anonymous 2006-01-09 3:34

there should be an answer. it makes a graph doesn't it? rotate it 90degrees and you have x=..

Name: Anonymous 2006-01-09 4:54

no. it's not solvable.

Name: Anonymous 2006-01-09 9:35

-[LambertW(-ln(a))]/ln(a)

Name: Anonymous 2006-01-09 11:12

elementary functions, even heard of them, you fruitcake? that's not a valid, closed-form solution.

Name: Anonymous 2006-01-09 11:53

let "1" + "1" = "2".
now find "2" in terms of "1".

Name: Anonymous 2006-01-09 12:46

>>23

Define an elementary function.

Name: Anonymous 2006-01-09 13:55

>>25

polynomials, trigonometric functions, hyperbolic functions, logarithmic functions, exponential functions, n-th root functions, and their FINITE, CLOSED-FORM combinations.

for instance, erf(x) and zeta(x) do not qualify.

Name: Anonymous 2006-01-09 17:57

The solution was already posted by >>4. Anyone posting after that is a fucking idiot.

Name: Anonymous 2006-01-09 19:48 (sage)

>>27

Especially you, for not knowing that 4channers don't care.

Name: Anonymous 2006-01-09 20:39

>>9

"obviously" lol, prove it fag.

Name: Anonymous 2006-01-10 11:50 (sage)

>>28
YOU JUST GOT OWNED AHAHAHAHAHAHAHAHAH FGT

Name: Anonymous 2006-01-10 16:31

needs more equations

Name: Anonymous 2006-01-11 5:41 (sage)

>>31
give a parametric solution then

Name: Anonymous 2006-01-11 15:15

>>32
y = x^(1/x)
http://en.wikipedia.org/wiki/Parametric

u do the rest

Name: Anonymous 2006-01-12 10:18

>>33
failt for learning math from wiki lol

Name: Anonymous 2006-01-22 16:50

>>2

Actually that's false. You can't derive an equation like this it's meaningless.

>>3
He's right.

Name: Anonymous 2006-01-23 20:55

make a proof

Name: Anonymous 2006-01-24 5:05 (sage)

even if I do a proof you won't understand it because you haven't mastered ZF, FOL or elementary analysis.

Name: Anonymous 2006-01-24 15:45

>>37
If you can't make a proof after spending 3 years studying maths in college or whatever it means you are stupid and inferior.

Name: Anonymous 2006-01-25 8:50

This equation doesn't have a definite solution without YOF or RV

Name: Anonymous 2006-01-25 12:34

>>38
I said that you're too dumb to understand it, not that I can't do it.

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