Solve x: Algebraically | x^3*e^(-a/x)=b

[Swapnil]

How would you solve for x algebraically?

$$x^3 e^{\frac{-a}{x}} = b$$

where a and b are some constants.

 

[berkeman]

Is this homework? If so, I can move it.

Start by isolating the logarithmic terms and the non-log terms. What can you do to get the x^3 away from the e^ term? Once you do that, what can you do to both sides of the equation to get rid of the e^?

 

[StatusX]

The Lambert W function is the first thing to try when you have something that looks like that. It has the property that W(x e^x) = x. So try to rearrange that into the form f(x) e^f(x) = C for some constant C, and then apply W to both sides to get f(x)=W(C).

 

[Swapnil]

Is this homework? If so, I can move it.

Start by isolating the logarithmic terms and the non-log terms. What can you do to get the x^3 away from the e^ term? Once you do that, what can you do to both sides of the equation to get rid of the e^?

I don't think that's going to help. If you do that, then you are just trapping x inside the natural log function instead of the exponential function.

 

[Swapnil]

The Lambert W function is the first thing to try when you have something that looks like that. It has the property that W(x e^x) = x. So try to rearrange that into the form f(x) e^f(x) = C for some constant C, and then apply W to both sides to get f(x)=W(C).

Wow! That's news to me. I searched the lambert W function on Wikipedia and I have to say it is pretty interesting. Let me see what I can do...

 

[Swapnil]

Got it!. I am 99% sure that the answer is:
$$x = \frac{a}{3W(\frac{ab^{-1/3}}{3})}$$.