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I have the following nonlinear equations which I wish to express in terms of Lambert W functions.

$$bx^a\exp(cx)=y\tag{1}$$

$$ (1+x^c)^{-m}bx^a=y\tag{2}$$

where $x$ is the root in both equations.

Is it possible to obtain closed form for the root $x$ in both equations using Lambert W function?

Thank you.

R.W
  • 2,504
S.A. Osagie
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  • Are they $bx^a e^{cx}=y$ and $(1+x^c)^{-m}bx^a=y$? For the first $x = \dfrac{a W\left(\dfrac{c \left(\dfrac{y}{b}\right)^{1/a}}{a}\right)}{c}$ – Moo Apr 03 '17 at 04:27
  • Thanks Moo. Yes they are. I appreciate – S.A. Osagie Apr 03 '17 at 05:32
  • As a rule of thumb, any equation which can write or rewrite $A+Bx+C\log(D+Ex)=0$ shows solution(s) in terms of Lambert function. As Moo commented, this is the case for the first equation. For the second, I do not think that it could be done. – Claude Leibovici Apr 03 '17 at 07:38
  • Thanks as well Claude. I will follow up and read more on Lambert W function. – S.A. Osagie Apr 03 '17 at 21:58
  • Are $a,b,c,m \neq 0$? In general what are the constraints of the parameters and variables? Are $x,y$ real? – Χpẘ May 04 '17 at 16:34
  • This paper, https://www.researchgate.net/publication/303839763_Fleshing_out_the_Generalized_Lambert_W_Function uses a technique in which the derivative is set to zero and then used to solve the original equation. If you do that for the second equation, then $x^c=\frac{a}{cm-a}$ and the equation can be inverted. – Χpẘ May 04 '17 at 16:47
  • Thanks Xpw for this information. It means I must take the derivative of the second equation before inverting to the Lambert W function. – S.A. Osagie May 07 '17 at 02:49

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