Math Help - Solving xe^x = 1

1. 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!

2. 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:

$x+1=1$

$x=0$

3. 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...

4. 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.

5. 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 $f'(x) = xe^x + e^x$

6. Look at the given function. If x = 0, then the whole left side becomes one. Anything to power zero is one.

7. Yes, but if x becomes 0, you have $0e^0 = 0$

8. x = ~.56714....

9. 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 $x_3$