# Solving equation

1. Mar 12, 2008

### hamamo

hi there;
plz can any one help me solving this
3((e^x)-1)-xe^x=0
sorry i couldn't use more elegant form to write the equation
i use some software and they help
but i cant do it in hand

2. Mar 12, 2008

### sutupidmath

$$3(e^{x}-1)-xe^{x}=0$$ I do not believe you can solve this one algebraically, one can only approximate the solution to these kind of equations.

3. Mar 12, 2008

### Marco_84

i think a graphical method is usefull...

ciao

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4. Mar 13, 2008

### hamamo

thanx for the reply up there
but i search more and i found this kind of equation can be solved using
Lambart w-function
or omega function, the problem i couldnt have more information about this function else some expansion series and i cant even write a code to solve or to find a value in lambart function
any more help will be useful
thanx

5. Mar 13, 2008

### hamamo

can you help me using the latex

6. Mar 13, 2008

### TheoMcCloskey

By inspection we can see x=0 is a solution. Do you have any reasoning to believe there are other solutions?

Edit = maybe I was to hasty - there seems that there is at least one more solution.

Last edited: Mar 13, 2008
7. Mar 13, 2008

### HallsofIvy

Staff Emeritus
If you click on the formula, you will see the code in a new window.

8. Mar 14, 2008

### hamamo

the 0 solution i know about it
and there is another solution if you graph the equation you can find it approximately

9. Mar 14, 2008

### hamamo

by the way
this equation is a result for the Blanck's low and Wien's displacment low
i want to calculate the Wien's constant at the maximum wave length of black body radiation
so
i differentiate Blanck's low and solve the equation for which x have a maximum value
and the result is something like this equation
which now i need to solve for x to find max and min value

10. Mar 14, 2008

### coomast

Hello Hamamo, if you want some code to calculate the Lambert W function, you might consider using the definition of it and the method of Newton-Raphson. The definition as you might know is:

$$X=Ye^Y \qquad \rightarrow \qquad Y=W(X)$$

Thus if you define a function f as:

$$f=Ye^Y-X$$

You can use the method of Newton Raphson to be for calculating this function:

$$Y_{n+1}=Y_n-\frac{Y_ne^{Y_n}-X}{e^{Y_n}(Y_n+1)}$$

Or:

$$Y_{n+1}=\frac{e^{Y_n}Y_n^2+X}{e^{Y_n}(Y_n+1)}$$

Take 0 as start value and use this iterative scheme to calculate the solution as the resulting value of the Lambert W function. It converges extremely fast. 5 iterations for the value of the function you are looking to solve.

best regards, Coomast

 The results of the iteration if you use it on your function:
step n Yn Yn+1
1 0 -0.149361
2 -0.149361 -0.177647
3 -0.177647 -0.178560
4 -0.178560 -0.178561
5 -0.178561 -0.178561

Which is x-3, thus x=2.821439 is the one you need

Last edited: Mar 14, 2008
11. Mar 14, 2008

### hamamo

thanx coomast
you r helpfull thats what i need
thanx again