Can you solve this?

[alkatran]

(X^X)^X = 3

Enjoy, my calculator gives me an error whenever I put it in. :D

[sql_lall]

Window Settings:
XMin=1
XMax=2
YMin=1
YMax=16 ((2^2)^2=16)
Graphing:
y=3
y=(x^x)^x

Intersection:
x=1.5655522757211(13 d.p)
y=3

[alkatran]

niiiiiiiiiiiiiiiice

Wtg, I'll have to remember how you did that ;)

[Guv]

[b]Sql_Lall[b] got it right. I verified his answer on my HP.

The derivative of (xx)x is a bit messy to determine, but you could use the Newton method to get the answer is you know the derivative.

[TheAlchemist]

correct me if im wrong but the derivative f(X) = (X^X)^X = X^X2

so if we look at it as a function of a function and solve by the chain rule then:
d/dx (f(X)) = X^2*(X^(X2 -1) * 2X

= (2X^(X2 +2)

[Guv]

TheAlchemist: (xx)x = xx^2 helps a lot. I feel stupid for not realizing that simplification.

Given Function(x) = xx^2, I do not think the derivative is as simple as your post.

I am not certain of the following, but think it is the way to approach this.

Function(x) = xx^2

Function(x) = [ eln(x) ]x^2, where ln(x) is natural log(x)

Function(x) = ex^2*ln(x)

Derivative[ eu ] = eu * Derivative(u)

Derivative[ ex^2*ln(x) ] = ex^2*ln(x) * Derivative[ x^2ln(x) ]

Now use derivative of a product.

[RAEsquivelC]

SQL_LLAL posted this answer to the x^x^x = 3 question:

x = 1.5655522757211(13 d.p)

By trial and error, using Excel, I got:
x = 1.56555227572112 (14 d.p., which, apparently, is the most I can get with Excel). It did take me some 5 minutes, though.

[TheAlchemist]

hmmm, yes Guv i see your point, its beautiful how you broke it down using e^ln(x). looking at it i now feel a bit dense that i applied methods, that ordinarily, would have been right but weren't so in this case because of the exponent X. dam why did't i see that!!!!