Solving xe^x = 1

• Nov 13th 2010, 10:45 AM
Yehia
Solving xe^x = 1
So I came across this equation:

xe^x=1

how would you solve that? there seems to be no analytical way of solving it. my teacher said something about iteration but any help would be appreciated. thanks!
• Nov 13th 2010, 10:50 AM
rtblue
Well, from intuition, we see that x=0 is a viable option.

$xe^x=1$

multiply both sides by xe.

$(xe^x)*(xe)=xe$

add the exponents on the left side, and we get:

$x+1=1$

$x=0$
• Nov 13th 2010, 10:57 AM
Yehia
Quote:

Originally Posted by rtblue
Well, from intuition, we see that x=0 is a viable option.

$xe^x=1$

multiply both sides by xe.

$(xe^x)*(xe)=xe$

add the exponents on the left side, and we get:

$x+1=1$

$x=0$

What? No idea how you went from step 2 to step 3.

and x cannot = 0 because then: 0=1 which we know cannot be true...
• Nov 13th 2010, 11:11 AM
FernandoRevilla
Quote:

Originally Posted by Yehia
So I came across this equation:
xe^x=1 how would you solve that? there seems to be no analytical way of solving it. my teacher said something about iteration but any help would be appreciated. thanks!

It is easy to prove that the equation has only one solution and this solution belongs to $(0,1)$. Now use for example the Newton-Raphson method.

Regards.
• Nov 13th 2010, 12:54 PM
Educated
I agree that x = 0 cannot be a solution.

Use the Lambert W function or an approximation method, like the Newton Raphson Method, to solve for x.

The Lambert W function has a simple solution, or if you plan to use the Newton Raphson method: $f(x) = xe^x -1$ then $f'(x) = xe^x + e^x$
• Nov 14th 2010, 07:58 AM
rtblue
Look at the given function. If x = 0, then the whole left side becomes one. Anything to power zero is one.
• Nov 14th 2010, 08:26 AM
Unknown008
Yes, but if x becomes 0, you have $0e^0 = 0$
• Nov 14th 2010, 09:09 AM
Wilmer
x = ~.56714....
• Nov 14th 2010, 11:18 AM
skeeter
Quote:

Originally Posted by Yehia
So I came across this equation:

xe^x=1

how would you solve that? there seems to be no analytical way of solving it. my teacher said something about iteration but any help would be appreciated. thanks!

Newton's method - Wikipedia, the free encyclopedia

$f(x) = xe^x - 1$

$f'(x) = e^x(x+1)$

$f(0) = -1$ , $f(1) = e-1$

since f(x) is continuous, there exists a zero between x = 0 and x = 1

Newton's method ...

$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}$

let $x_0 = 0.5$

$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 0.571020439808...$

$x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 0.567155568744...$

$x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = 0.567143290533...$

continued iterations will be extremely close to $x_3$