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I am trying to use the Lambert-W function to solve the following equation for $x$.

$$ \frac{a}{b} \ln x-x+ c =0 $$

My first step is to rewrite $x$ as $e^{\ln x}$.

$$ \frac a b \ln x - e^{ln x} + c =0$$

I then multiply by $b$ and divide by $a$.

$$ \ln x - \frac b a e^{\ln x} + \frac b a c =0 $$

I have read about the Lambert-W function but am unsure of how to progress!

Leucippus
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RedPen
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1 Answers1

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Write your equation as $$ \ln(x) - \frac{bx}{a} = - \frac{bc}{a} $$ take the exponential of both sides: $$ x e^{-bx/a} = e^{-bc/a}$$ and multiply by $-b/a$. With $u = -bx/a$ we have $$ u e^u = -\frac{b}{a} e^{-bc/a} $$ Thus $u = W\left(-\frac{b}{a} e^{-bc/a}\right)$, and $$ x = -\frac{a}{b} W\left(-\frac{b}{a} e^{-bc/a}\right) $$

Robert Israel
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  • When everything is complex numbers. Is there a way to obtain a series representation of the value of W at that point, is it possible to obtain where is the range of W at that point? I am particularly interesting in knowing if the solutions of the initial equation could be near the real axis – Tom Builder May 04 '20 at 23:35
  • There are series representations of $W$. For example, $$W(z) = \sum_{k=1}^\infty \frac{(-k)^{k-1} z^k}{k!}$$ for $|z| < 1/e$. – Robert Israel May 05 '20 at 02:35