I need to solve the next equation:


where a, b, c and d are positive real values.

Do I need to use Lambert W function, or there is some other method?


  • $\begingroup$ WLOG, assume $a = 1$. (why ?). You will need some form of LambertW or other non-elementary function to express $x$. $\endgroup$ – AlexR Aug 26 '13 at 9:27
  • $\begingroup$ Yeah, not problem for a, but problem is with constant c in exponent. I don't know how to solve it using Lambert W, could you be more precise, or share me some link with similar examples? Thanks! $\endgroup$ – Ivan Aug 26 '13 at 9:36
  • $\begingroup$ en.wikipedia.org/wiki/Lambert_W_function check examples here. Your question is not a standard one. $\endgroup$ – Seyhmus Güngören Aug 26 '13 at 9:39
  • $\begingroup$ Sorry, can't really help you on that right now - don't have Mathematica / Maple at hand and W|A computation time exceeds ^^ I tried substituting $u = bx+c$ but that just shifts the problem into the logarithm... $\endgroup$ – AlexR Aug 26 '13 at 9:43

The equation $ax^{bx+c}=d$ will not have its solution expressed in terms of lmabert function.

To make the story short, you can consider that, if the equation can be rewritten as $$A+Bx+C\log(Dx+E)=0$$ its solution will be a Lambert function. $$x=\frac{C}{B}W\left(\frac{B e^{\frac{B E-A D}{C D}}}{C D}\right)-\frac{E}{D}$$

For $ax^{bx+c}=d$, the problem is different since I think that any rearrangement of the equation would include an extra term looking as $x\log(x)$


$$ax^{bx+c}=d$$$$ae^{(bx+c)\ln(x)}=d$$$$x=W(u)$$$$e^{(bW(u)+c)\ln(W(u))}=\frac da$$$$\ln(W(u))=\ln(u)-W(u)$$$$(bW(u)+c)(\ln(u)-W(u))=\ln(\frac da)$$$$-bW^2(u)+(b\ln(u)-c)W(u)+c\ln(u)-\ln(\frac da)=0$$$$W(u)=\frac{c-b\ln(u)\pm\sqrt{(b\ln(u)-c)^2+4b(c\ln(u)-\ln(\frac da))}}{-2b}=\frac{c-b\ln(u)\pm\sqrt{b^2\ln^2(u)-2bc\ln(u)+c^2+4bc\ln(u)-4b\ln(\frac da)}}{-2b}=\frac{c-b\ln(u)\pm\sqrt{b^2\ln^2(u)+2bc\ln(u)-4b\ln(\frac da)+c^2}}{-2b}$$$$u=xe^x$$$$x=\frac{c-b\ln(xe^x)\pm\sqrt{b^2\ln^2(xe^x)+2bc\ln(xe^x)-4b\ln(\frac da)+c^2}}{-2b}$$$$=\frac{c-b[\ln(x)+x]\pm\sqrt{b^2[\ln(x)+x]^2+2bc[\ln(x)+x]-4b\ln(\frac da)+c^2}}{-2b}$$$$x=\frac{c-b[\ln(x)+x]\pm\sqrt{b^2\ln^2(x)+2b^2x\ln(x)+b^2x^2+2bc\ln(x)+2bcx-4b\ln(\frac da)+c^2}}{-2b}$$

I'm sorry to disappoint, but the c and d in the end of the square root will make it impossible to have the square root of that come out and is therefore unsolvable.

Moving things to the left and squaring both sides won't work either because you will simply get some true statement or an earlier step.

I have, however, heard of manipulating square roots somehow and wonder if that could help...


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