How to solve exponential equation of $x + e^{-x }= 3$? I've got the following equation: 
$\lambda + e^{-\lambda} = 3$
$3 - \lambda  = \frac{1}{e^{\lambda}}$
$\frac{1}{3-\lambda} = e^{\lambda}$
Now, I take the natural log of both sides: 
$ln(\frac{1}{3-\lambda}) = \lambda$
$ln(1) - ln(3-\lambda) = \lambda$
Since $ln(1)$ is zero,
$-ln(3 - \lambda) = \lambda$
I get to this part of the equation and I feel stuck and that I cannot simplify and solve for lambda even more. Could someone help me? Obviously I can use a graphing calculator to find the solution (2.9475309 I believe being one of them), but I would like to learn how to solve this by hand as well. Thank you in advance. 
 A: Newton-Raphson method may be handy here for finding numerical solutions. Define $f(x)=x+e^{-x}-3$. Notice that we're actually solving $f(x)=0$.
$$x_{n+1}=x_n-\dfrac{x_n+e^{-x_n}-3}{1-e^{-x_n}} \land x_0=3$$
$$\begin{array}{|p{3cm}||p{3cm}|p{3cm}|p{3cm}|}\hline n & x_n  \\ \hline 0 & 3 \\ 1 &  2.94760430351\\ 2& 2.94753090269\\3&2.94753090254\\\hline \end{array}$$
There is also a negative solution existence of which somehow makes sense because of the $e^{-x}$ term. As $x$ decreases, $e^{-x}$ increases. In fact it calls for the existence of a value such that $x$ decreases and $e^{-x}$ increases in such a way that they nullify the $+3$ term of the equation providing a negative solution.
Let $x'$ denote the second solution. Again using the first iteration formula but now with $x'_0=-3$. We get the following iterations: $$\begin{array}{|p{3cm}||p{3cm}|p{3cm}|p{3cm}|}\hline n & x'_n  \\ \hline 0 & -3 \\ 1 &  −2.26197848246\\ 2& −1.75743824023\\3&−1.54063495725\\\hline \end{array}$$

Aliter:
We can do a substitution of variables to make it satisfy being of the form wherein we can use Lambert's $W$ function.
Let $-t=x-3$. Now we have $tp^t=e^3$ which has solution $t=-W(-e^3)$ and so $x$ is $3-t$ or simply $x=3+W(-e^3)$.
A: Considering that you look for the zero's of
$$f(x)=x+e^{-x}-3$$ notice that the fist derivative cancels at $x=0$. To have an idea about the location of the roots, perform a Taylor expansion to get
$$f(x)=-2+\frac{x^2}{2}+O\left(x^3\right)$$ Ignoring higher order terms, this gives as approximate solutions $x_{\pm}=\pm 2$ that you can use as starting guesses for Newton method
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 2.00000 \\
 1 & 3.00000 \\
 2 & 2.94760 \\
 3 & 2.94753
\end{array}
\right)$$
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & -2.00000 \\
 1 & -1.62607 \\
 2 & -1.51397 \\
 3 & -1.50529 \\
 4 & -1.50524
\end{array}
\right)$$
You have much better guesses building around $x=0$ the $[2,n]$ Padé approximant of the function. It will write
$$f(x) \sim \frac{-2+a_1^{(n)}x+a_2^{(n)}x^2}{1+\sum_{k=1}^n b_k x^k}$$ and then solve the numerator (simple).
As a function of $n$, the following table gives the decimal representation of the guesses
$$\left(
\begin{array}{ccc}
n & x_-^{(n)}& x_+^{(n)} \\
 2 & -1.50000& 3.00000 \\
 3 & -1.50624& 2.95069 \\
 4 & -1.50554& 2.93996 \\
 5 & -1.50524& 2.94888 \\
 6 & -1.50523& 2.94834
\end{array}
\right)$$
A: This is an example of equation, for which
we know there is 
an explicit solution 
in terms of
well-known
Lambert $\operatorname{W}$ function.
\begin{align} 
x + \exp(-x)&= 3
\tag{1}\label{1}
,\\
x\exp(x) + 1&= 3\exp(x)
,\\
x\exp(x)-3\exp(x)&=-1
,\\
(x-3)\exp(x)&=-1
,\\
(x-3)\exp(x-3+3)&=-1
,\\
(x-3)\exp(x-3)\exp(3)&=-1
,\\
(x-3)\exp(x-3)&=-\exp(-3)
.
\end{align} 
The last expression is ready to apply
the Lambert $\operatorname{W}$ function:
\begin{align} 
\operatorname{W}((x-3)\exp(x-3))&=\operatorname{W}(-\exp(-3))
,\\
x-3&=\operatorname{W}(-\exp(-3))
,\\
x&=3+\operatorname{W}(-\exp(-3))
\tag{2}\label{2}
.
\end{align} 
Equation \eqref{2} provides
a general solution in terms of 
Lambert $\operatorname{W}$ function,
and since we are interested in real solutions,
we need to check the argument of $\operatorname{W}$.
In this case it is $-\exp(-3)$, which is 
between $-\exp(-1)$ and $0$. 
This condition guarantees that there are 
two real solutions, corresponding to 
so-called branches 
of the Lambert $\operatorname{W}$ function,
which has special names 
$\operatorname{W_{0}}$ and
$\operatorname{W_{-1}}$:
\begin{align}
x_0&=3+\operatorname{W_{0}}(-\exp(-3))
\approx \phantom{-}2.94753090254
,\\
x_{-1}&=3+\operatorname{W_{-1}}(-\exp(-3))
\approx -1.50524149579
.
\end{align}
