How to solve $2x-e^{-x}=0$? I am wondering how to solve the equation $2x-e^{-x}=0$?
I tried:
$2x=e^{-x}\implies\log(2x)=-x$
$\implies\log2+\log x=-x\implies\log2=-x-\log x$
Then I am stuck.
Is there any elementary way to solve the equation?
Thanks.
 A: $$
2x - e^{-x} = 0 \iff \\
2x\,e^x -1 = 0 \iff \\
x \, e^x = 1/2 \iff \\
x = W(1/2) = 0.3517337\dotsb
$$
where $W$ is the inverse to $f(x) = x e^x$. This is known as Lambert W function and is a special function, not an elementary one.
A: 
This is what I mean.$$2x=g(x)\\f(x)=e^{-x}$$plot both of them .the cross section point is solution of equation. 
  we can approximate it by numerical methods .
A: Tou can't get result in closed form, but there are some easy possibilities to calculate it:


*

*To calc W(1/2) with Wolfram Alpha

*To use series from the article

*To use iterative method
$$x_{i+1}=\dfrac{\exp(-x_i)+x_i}3$$
with $x_0=0.$
A: If you do not (or cannot) use Lambert function, numerical mathods should be used.
Probably the simplest could be Newton method which, starting from a guess $x_0$, will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ For the case where $$f(x)=2x-e^{-x}\implies f'(x)=2+e^{-x}\implies x_{n+1}=\frac{x_n+1}{2 e^{x_n}+1}$$ Being lazy, let us start using $x_0=0$. Then the sucessive iterates will be $$x_1=0.3333333333$$ $$x_2=0.3516893316$$ $$x_3=0.3517337110$$ $$x_4=0.3517337112$$ which is the solution for ten significant figures.
