Solve for x in $3(1-e^{-x})=x$. $3(1-e^{-x})=x$ . We nees to find the value of $x$ from the equation.

I have tried using expansion of exp but thats not fruitful....any ideas are welcome...

Thank You...
$$x-3=-3e^{-x}$$
$$(x-3)e^x=-3$$
$$(x-3)e^{x-3}= -3e^{-3}$$
$\begin{cases} X=x-3 \\ Y=-3e^{-3} \end{cases} \quad\to\quad Xe^X=Y \quad\to\quad X=W(Y)$

$W$ is the Lambert's W function.
$$x-3=W\left(-3e^{-3}\right)$$
$$x=3+W\left(-3e^{-3}\right)$$
The Lambert's W function is a multi valuated function :

http://mathworld.wolfram.com/LambertW-Function.html 

First real value : $W\left(-3e^{-3}\right)=-3$ and second real value : $W\left(-3e^{-3}\right)\simeq -0.17856$

The real solutions are : $x=0$ and $x\simeq 2.82144$

In practice, there is no need for Lambert-W function. One can directly compute an approximate of the root thanks to numerical methods, for example Newton-Raphson.
$e^{-x}=1-\frac {x}{3}$

Let $y = 1-\frac {x}{3} \Rightarrow x=3(1-y)$

$e^{-3(1-y)}=y$

$e^3 \cdot e^{-3y}=\frac {1}{y}$

$e^{-3y}=\frac {e^{-3}}{y}$

Powering both sides by $\frac {e^{-3}}{y}$ to eliminate $y$ in the left side

$e^{-3e^{-3}}={\frac {e^{-3}}{y}}^{\frac {e^{-3}}{y}}$

Base and power are equal for both sides $a^a=z^z$, so one of the solutions is to let

$e^{-3}=\frac {e^{-3}}{y} \Rightarrow y=\frac {e^{-3}}{e^{-3}}=1$

then $x=3(1-y) =0$

Note that the super square root of $a^a$ may has 2 real values.

for example $z^z = {\frac {1}{2}}^{\frac {1}{2}}$ has 2 real solution
$z= \frac{1}{2}$ and $z= \frac {1}{4}$