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How do I solve the following equation? I understand it is not possible with regular arithmetic steps:

$$xb^x = a$$

Provided $b \in \mathbb{R}^+$ and $a \in \Bbb{R}$. What would be the formal explanation telling this cannot be solved by regular arithmetics, and what would be a good way to solve it?

Peter Woolfitt
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Luis Masuelli
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    Just a nitpick - this can't be solved in general, since it clearly can be solved for some specific choices of $a,b$. Also, would you be so kind as to specify what you mean by "regular arithmetic"? – Wojowu Nov 27 '15 at 15:28
  • related: http://mathoverflow.net/questions/135911/how-to-prove-lamberts-w-function-is-not-elementary – Myself Nov 27 '15 at 15:30
  • Regular arithmetic steps to solve an equation (i.e. apply the same operations -except mul/div by zero- at both sides of the equation) – Luis Masuelli Nov 27 '15 at 15:33
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    Would using logarithms be allowed as that may be a way to get an isolation of x? – JB King Nov 27 '15 at 15:39
  • with the so-called Lambert we get $$x={\frac {{\rm W} \left(\ln \left( b \right) a\right)}{\ln \left( b \right) }} $$ – Dr. Sonnhard Graubner Nov 27 '15 at 16:00

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Write the equation as $$ x\,e^{(\log b)x}=a. $$ Multiply the equation by $\log b$: $$ (\log b)x\,e^{(\log b)x}=a\log b. $$ The Lambert function $W(z)$ is defined as the inverse of $z\,e^z$: $$ (\log b)x=W(a\log b). $$ Finally $$ x=\frac{W(a\log b)}{\log b}. $$

Julián Aguirre
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  • Aaaah but does $W$ have an algebraic expression somehow? Or is just defined like that (as $Erf$ is defined as the antiderivative of $e^{-x^2}$)? – Luis Masuelli Nov 27 '15 at 17:00
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    $W$ is a trascendental function. Unlike the error function, it is elementary, since it is the inverse of an elementary function. – Julián Aguirre Nov 27 '15 at 17:02