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...
 A: $$x-3=-3e^{-x}$$
$$(x-3)e^x=-3$$
$$(x-3)e^{x-3}= -3e^{-3}$$
$\begin{cases} X=x-3 \\ Y=-3e^{-3} \end{cases} \quad\to\quad Xe^X=Y \quad\to\quad X=W(Y)$
$W$ 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. 
A: Maybe your equation is better to discuss, if you check the characteristics of $y=x^{1/x}$, because the equation is equivalent to $${x_1}^{1/{x_1}}={x_2}^{1/{x_2}}$$ with $\displaystyle x_1:=e^3(1-\frac{x}{3})$ and $x_2:=e^3$.
It is $x=3(1-e^{-3}x_1)$.
Then you see, that $x_1$ (and therefore $x$) has two solutions: 
(1) $x_1>e$: $x_1=x_2$ which means $x=0$ 
(2) $1<x_1<e$:
The recursion $$a_{n+1}=({x_2}^{1/{x_2}})^{a_n}$$ with the start value $a_0:=1$ gives you the second solution $x_1=\lim_{n\to \infty}a_n$.
A: You can't get the exact solution of this equation, so you need an approximation. But it's good practice to first know how many solutions there may be, which you can figure out (or at least guess) if you sketched the graphs of the LHS and RHS.
First, note that $x=0$ is a solution, and that the derivative of $3(1-e^{-x})$ is $3e^{-x}$ and decreasing. At $x=0$, the value of this derivative is 3, which is larger than the derivative of $x$ (the RHS, the derivative of which is 1 everywhere). However, when $x$ gets large, the LHS derivative approaches 0, which is less than the RHS derivative. Taken together with the fact that both the LHS and RHS are increasing in $x$, you can see that there must be a second point where these functions cross. On the other hand, there is no solution for $x<0$, as $3e^{-x} < x$ for all $x<0$ (which you can argue by a similar method if necessary).
So you now know that there is only one other solution, and it's positive; you can either get this with a numerical method (say Newton-Raphson), or by taking the Taylor expansion of $e^{-x}$ and solving the arising polynomial.
A: $$ y= 3(1-e^{-x}) -x $$
$x=0$ is a solution by inspection. The other is a root that can be found exactly only by a numerical method such as Newton-Raphson or Regula Falsi. Draw a graph of the above transcendental function to find its approximate location near about $ x \approx 2.8 $
A: The function $f(x)=3-3e^{-x}-x$ has


*

*$\displaystyle\lim_{x\to-\infty}f(x)=-\infty$

*$\displaystyle\lim_{x\to\infty}f(x)=-\infty$

*$f'(x)=3e^{-x}-1$
The derivative vanishes only at $x=\log3$; it's easy to check that the function is increasing in the interval $(-\infty,\log3]$ and decreasing in the interval $[\log3,\infty)$. Since
$$
f(\log3)=3-3e^{-\log3}-\log3=2-\log3>0
$$
the equation $f(x)=0$ admits two solutions. One solution is $0$ (the one in $(-\infty,\log3)$, and the other one is $>\log3$.
There is no “explicit” form of the solution, but you can approximate it with numerical methods.
A: HINT.-Without Lambert-W function, you have to apply numerical approximation which can be done different ways. For example, consider the function $$g(x)=3(1-e^{-x})$$ It is clear that $g(0)=0$ so $x=0$ is a (trivial) solution, besides $g(x)\lt 3$ for all real $x$. Furthermore $g(x)$ is a concave downward function hence there is just one other solution which is a fixed point of $g$.
We have $$g(2)\approx 2.5\gt2\\g(3)\approx 2.85\lt3$$
It follows $g(x_0)=x_0$ for some unique $x_0$ such that $2\lt x_0\lt3$.
We follow with $g(2.6)\approx 2.7\gt 2.6$ so $2.6\lt x_0\lt3$ and so on we can refined the approximation till $g(x)\approx x$
A: $e^{-x}=1-\frac {x}{3}$
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 $z= \frac {1}{4}$
