Exponential equation(Solution) I'm solving one passion distribution question and I face this algebra problem which makes me can't proceed.How to solve the equation $$e^{-x}(1+x)=0.8$$?
 A: You can write it as $(10+10x)=8e^x$ and then use taylor series of $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}...$ and find approximate value of $x$
A: If you are willing to use the Lambert-W Function we get the following result
$$e^{-x}(1+x)=\frac{4}{5}$$
$$(-x-1)e^{-x-1}=-\frac{4}{5e}$$
$$-x-1=\operatorname{W}_n\left(-\frac{4}{5e}\right)$$
$$x=-\operatorname{W}_n\left(-\frac{4}{5e}\right)-1$$  
Note that this yields two real solutions, namely
$$x=-\operatorname{W}_0\left(-\frac{4}{5e}\right)-1 \approx -0.528328$$
$$x=-\operatorname{W}_{-1}\left(-\frac{4}{5e}\right)-1 \approx 0.824388$$
There are also a number of complex solutions; these are encapsulated by using the $n$th branch of the W function
A: Fun with Math time
Just to let you to solve that equation in a more amusing way, without strange functions! Please: note that this is just a numerical way to obtain the solutions, so it's less precise than the other answers, but you'll see is not that bad!
First just call $x = \ln(a+1)$, and substitute:
$$e^{-\ln(a+1)}(1 + \ln(a+1)) = 0.8$$
namely
$$\frac{1}{a+1}(1 + \ln(a+1)) = 0.8$$
Now one could use the logarithm series, up to the second order:
$$\ln(a+1) \approx a - \frac{a^2}{2}$$
and substituting you get
$$\frac{1}{a+1}\left(1 + a - \frac{a^2}{2}\right) = 0.8$$
Now you multiply by $a+1$ both sides and arranging the terms you'll end up with a quadratic equation:
$$a^2 - 0.4a - 0.4 = 0$$
Using the well known formula for second degree equations, your solutions will be
$$\begin{cases}
a_1 \approx 0.86332495 \\
a_2 \approx -0.46332495
\end{cases}$$
As I said: not that precise, but you have a very good idea about the solutions. 
Final Remark
This is the way in which you may work if you have no calculator near you! 
A: Equations which mix polynomial, exponential and/or trigonometric functions do not show explicit solutions and numerical methods are required.
As Carl Heckman commented, the equation $$e^{-x}(1+x)=a$$ has a solution in terms of Lambert function. It is sufficient to rewrite it as $$e^{-(1+x)}(1+x)=\frac a e$$ and the solution write $$x=-W\left(-\frac{a}{e}\right)-1$$ where $W(z)$ is Lambert function. The Wikipedia page gives expansion formulae for its evaluation. 
In the case of $a=0.8$, there are two solutions $x_1\approx -0.528328$ and $x_2\approx 0.824388$.
Otherwise, Newton method is a simple way to go.
A: You can solve this numerically using the Newton-Raphson method.
$$\begin{align}
y & = e^{-x} (1 + x)\\
\frac{dy}{dx} & = e^{-x}  - e^{-x} (1 + x)\\
& = e^{-x}(1  - 1 - x)\\
& = -xe^{-x}\\
\Delta x & = \frac{\Delta y}{dy/dx}\\
& = \frac{k - e^{-x}(1 + x)}{-xe^{-x}}\\
\Delta x & = \frac{1 + x - ke^x}{x}
\end{align}$$
Here's a short Python program that uses that formula for $\Delta x$ to solve the equation in the OP. It gets 15 decimal places of accuracy in 4 loops.
from math import exp

def solve(x, k):
    while True:
        dx = (1.0 + x - k * exp(x)) / x
        x += dx
        if abs(dx) < 1.0E-15:
            break
    return x

x = 1.0
x = solve(x, 0.8)
print('%.15f %.15f' % (x, exp(-x) * (1 + x)))

output
0.824388309032984 0.800000000000000

And here's a more accurate approximation, calculated using the same algorithm in mpmath:
0.82438830903298460937957300776559646539962691434071848356636393644902371149509543402742674738895867255235545074861413882805277214780192444340
