Using the Lambert W-Function to Solve for a Unique Real Value of x

[Uan]

Not really sure where this question belongs in this forum...

I was solving an engineering problem and I got to the form

ax=b-cln(dx)

where a, b, c and d are constant real values. I had a peek at the answer and they got a unique positive real valued answer for x but I have no idea how. Some searching I came across the Lambert W-Function and I got it into the form

\frac{1}{d}e^{\frac{b}{c}} = xe^{\frac{ax}{c}}

How do I proceed to apply the Lambert W-Function from here?

WolframAlpha found that

x = \frac{c}{a}W\left ( \frac{a}{cd}e^{\frac{b}{c}} \right )

 

[mfb]

$$x=W(g) \Leftrightarrow x*e^x = g$$
Therefore,
$$\frac{ax}{c} = W\left ( \frac{a}{cd}e^{\frac{b}{c}} \right ) \Leftrightarrow \frac{ax}{c} \exp\left(\frac{ax}{c}\right) = \frac{a}{cd}e^{\frac{b}{c}}$$

Multiply both sides with c/a and you get the equation in your post. The other direction works the same, you just have to see that you need the shape (something)*e^(something) and work with the constants to get that.

 

[ShayanJ]

Well, LambertW is defined to be the inverse of W e^W. If there was a way to invert that function, there would be no need to define LambertW! The expression you got is the most thing you can do. Then you are supposed to say "Oh yeah, so x should be the LambertW of such and so". And for actually finding numbers for x, you should consult mathematical tables or math softwares .

 

[Uan]

Thanks, that clears a lot up. So really the Lambert W in this case doesn't help all that much - just allows the function to be in a more recognisable form. You still need to go back to...

\frac{1}{d}e^{\frac{b}{c}} = xe^{\frac{ax}{c}}

and solve numerically for x.

 

[mfb]

Or look up function values of the Lambert W function, yes.