Question

How do I solve \(y=x^x\) for \(x\)?

Answer 1

that is not the answer

Answer 2

you're going to maybe use logarithmic on this one

Answer 3

I can't get past \(\ln{y} = x\ln{x}\) if I do that.

Answer 4

take the derivative of x ln (x)

Answer 5

This isn't a limit, I don't see any justification for taking a derivative of either side.

Answer 6

if you take the derivative, you can isolate y
and you already have y' in terms of x from calculus

Answer 7

@ganeshie8 @wio
summoning the demi-gods of math

Answer 8

y= y'/(ln(x)+1)

Answer 9

How would that find \(x\) though?

Answer 10

whoops sorry...wrong way

Answer 11

try its roots, it might reveal something

Answer 12

wats wrong in using linear approximation/newton's method as suggested by celestial/oidbio ?

Answer 13

x= e^(y'/y - 1)

Answer 14

and replace y with x^x
and replace y' with the derivative of x^x

Answer 15

@ganeshie8 The goal was to solve for \(x\), not approximate particular values.

Answer 16

you can't solve for a specific value of x because they are place holder
the closest you can get to a numerical value is the euler's constant

Answer 17

y' = (ln(x)+1)*x^x

Answer 18

@celestialdictator By solve for \(x\) I meant isolate \(x\).

Answer 19

then we will have to define a new function :
http://en.wikipedia.org/wiki/Lambert_W_function

Answer 20

Yeah...no

Answer 21

well x is a function of y and y'

Answer 22

Well that's interesting.

Answer 23

check example 3 of what ganeshie posted

Answer 24

you can have an infinite x^x^x^x^x...

Answer 25

yeah im also seeing these for the first time...

x^x= y


\(x = e^{W(ln y)}\)

Answer 26

then proceed with example 4, which what you were shown just now

Answer 27

look at 2nd reply
http://math.stackexchange.com/questions/50316/xx-y-how-to-solve-for-x

Answer 28

I'm surprised that this can't be handled with normal functions given how simple it seems.

Answer 29

It's not that simple because x is operating on itself so any normal inverse would involve x as well.