How to solve when x is an exponent and a base

[JungleJesus]

Just out of curiosity, I was wondering how to solve for x in an expression of the form
a^x = x^b

It kinda tripped me out at first.

 

[nicksauce]

There is no way in general to solve this kind of equation in terms of elementary functions.

 

[Anonymous217]

You can try manipulating it to fit the lambert W function I think.

 

[HallsofIvy]

You can try manipulating it to fit the lambert W function I think.

Yes. Taking the "b"th root of both sides, (a^{1/b})^x= x so that 1= xe^{-(1/b)(ln a)x}. Multiply on both sides by -(1/b)ln a: -(1/b)ln a= (-(1/b)(ln a)x)e^{-(1/b)(ln a)x}.

Letting y= -(1/b)(ln a)x, that becomes ye^y= -(1/b)ln a. Using the Lambert W function (which is defined as the inverse function to f(x)= xe^x) we have y= W(-(1/b)ln a).

Then x= -\frac{b}{ln a}W(-(1/b)ln a)

 

[JungleJesus]

Letting y= -(1/b)(ln a)x, that becomes ye^y= -(1/b)ln a. Using the Lambert W function (which is defined as the inverse function to f(x)= xe^x) we have y= W(-(1/b)ln a).

Then x= -\frac{b}{ln a}W(-(1/b)ln a)

Is there any way to simplify this expression, or is this the bare bones? How would one procede to calculate this value?

 

[CRGreathouse]

Is there any way to simplify this expression, or is this the bare bones? How would one procede to calculate this value?

There are numerical methods, like Newton's method or the secant method, that can give numerical values for the W function -- or even your original function directly. (If you just want the W function, there are optimizations that can be used to make it go faster.)