Lambert W function? I need to solve the next equation:
$$ax^{bx+c}=d$$
where a, b, c and d are positive real values.
Do I need to use Lambert W function, or there is some other method? 
Thanks!
 A: 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 $x\log(x)$ 
A: $$ax^{bx+c}=d$$$$ae^{(bx+c)\ln(x)}=d$$$$x=W(u)$$$$e^{(bW(u)+c)\ln(W(u))}=\frac da$$$$\ln(W(u))=\ln(u)-W(u)$$$$(bW(u)+c)(\ln(u)-W(u))=\ln(\frac da)$$$$-bW^2(u)+(b\ln(u)-c)W(u)+c\ln(u)-\ln(\frac da)=0$$$$W(u)=\frac{c-b\ln(u)\pm\sqrt{(b\ln(u)-c)^2+4b(c\ln(u)-\ln(\frac da))}}{-2b}=\frac{c-b\ln(u)\pm\sqrt{b^2\ln^2(u)-2bc\ln(u)+c^2+4bc\ln(u)-4b\ln(\frac da)}}{-2b}=\frac{c-b\ln(u)\pm\sqrt{b^2\ln^2(u)+2bc\ln(u)-4b\ln(\frac da)+c^2}}{-2b}$$$$u=xe^x$$$$x=\frac{c-b\ln(xe^x)\pm\sqrt{b^2\ln^2(xe^x)+2bc\ln(xe^x)-4b\ln(\frac da)+c^2}}{-2b}$$$$=\frac{c-b[\ln(x)+x]\pm\sqrt{b^2[\ln(x)+x]^2+2bc[\ln(x)+x]-4b\ln(\frac da)+c^2}}{-2b}$$$$x=\frac{c-b[\ln(x)+x]\pm\sqrt{b^2\ln^2(x)+2b^2x\ln(x)+b^2x^2+2bc\ln(x)+2bcx-4b\ln(\frac da)+c^2}}{-2b}$$
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...
