solving exponential using Quadratic

Question

How should i solve $\ e^{-x}(x+1)=0.8$ by Quadratic? I don't understand how should we make it linear, i tried to expand it and it is always the extra x beside the exp(-x) that make me struggle.

Answer 1

As Eevee Trainer commented, welcome to the world of Lambert function !

Just rewrite $$ e^{-x}(x+1)=a \implies e^{-(x+1)}(x+1)=\frac a e$$ to get, as shown in the Wikipedia page, $$x=-W\left(-\frac{a}{e}\right)-1$$ For a numerical evaluation in your case, since the argument is close to $-\frac 1e$, use the Taylor series $$W(t)=-1+\sqrt{2 e} z-\frac{2 e }{3}z^2+\frac{11 e^{3/2} }{18 \sqrt{2}}z^3+\cdots \qquad \text{where} \qquad z=\sqrt{t+\frac 1e}$$ You would get $x\approx -0.538$ while the exact solution is $-0.528$.

As said, numerical methods have to be used in order to polish the root. The Wikipedia page tells about the Halley method used by Corless.