Given a>0 solve the equation:

 

x^x[a(a+1)]) = a.

To be honest, I think all the superscripts in the problem here is making it harder to read rather than easier. Here is the problem stated wihout the use of superscript:

 

x^x^[a^(a+1)] = a.

 

I dont know. Use whichever seems easiest to look at.

WHOA!!! I have never seen so many indices ever in my life! Where did u get THAT from??!

How did you make that first one?

 

By the way, does anyone know good math forum?

a=1 and x=1 ,seems to work fine

ln a * e^-(a^(a+1)) = x + ln x

Actually to solve x it would go:

 

x = (a^a+1Log(a)/Productlog[a^a+1Log(a)])^a^a-1

Actually to solve x it would go:

 

x = (a^a+1Log(a)/Productlog[a^a+1Log(a)])^a^a-1

 

To be quite honest, I'm not sure what Productlog is. How does it relate ot Log or nl? And how did you arrive at this answer?

 

Suppose that a = 1/2. What would Productlog[a^a+1Log(a)] give as a result?

ProductLog is another name for the Lambert W relation.

wow, this problem seem to have some problem. "x^" cannot be followed by "[a^(a+1)]" I'm surprised, not even my super powerful mathematica 5.1 seems to slove it.

Could you show your work to get to the answer?

 

I'm stuck at

 

x * ln x = ln(a) / (a+1)

i think it would be preferable using LATEX.

 

This is what i get from banging in equation in Maple

 

[math]\exp[{\frac{\mbox{LambertW}(\ln(a)a^{a+1})}{a^{a+1}}][/math]

 

to dimshadow. maple gives this

 

[math]\frac{\ln(a)}{(a+1)\mbox{LambertW}(\frac{ln(a)}{a+1})}[/math]

 

i dont know LW function personally, but u can have a read about it at

 

http://mathworld.wolfram.com/LambertW-Function.html