- #1
Swapnil
- 459
- 6
How would you solve for x algebraically?
[tex]x^3 e^{\frac{-a}{x}} = b [/tex]
where a and b are some constants.
[tex]x^3 e^{\frac{-a}{x}} = b [/tex]
where a and b are some constants.
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.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^?
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...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).
The equation is x^3*e^(-a/x)=b.
x represents the unknown variable we are trying to solve for, a represents a constant, and b represents a known value or constant.
Yes, this equation can be solved algebraically by manipulating the terms and isolating the variable x.
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.
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.