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

• Swapnil
In summary, to solve for x algebraically in the equation x^3 e^{\frac{-a}{x}} = b, one can use the Lambert W function, which has the property W(x e^x) = x. By rearranging the equation into the form f(x) e^f(x) = C and applying W to both sides, the solution for x is x = \frac{a}{3W(\frac{ab^{-1/3}}{3})}.
Swapnil
How would you solve for x algebraically?

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

where a and b are some constants.

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^?

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).

berkeman said:
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.

StatusX said:
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...

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

Last edited:

## What is the equation for solving x?

The equation is x^3*e^(-a/x)=b.

## What do the variables in the equation represent?

x represents the unknown variable we are trying to solve for, a represents a constant, and b represents a known value or constant.

## Can this equation be solved algebraically?

Yes, this equation can be solved algebraically by manipulating the terms and isolating the variable x.

## What is the purpose of using e in the equation?

e is a mathematical constant, also known as Euler's number, which is approximately equal to 2.718. It is commonly used in exponential functions and allows us to model natural growth or decay.

## Are there any real solutions to this equation?

Yes, there may be real solutions to this equation depending on the values of a and b. The solutions can be found by using algebraic techniques such as factoring, substitution, or the quadratic formula.

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