Solving xe^x = 1

**Yehia** · November 13th 2010, 09:45 AM

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!

**rtblue** · November 13th 2010, 09:50 AM

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:

$x+1=1$

$x=0$

**Yehia** · November 13th 2010, 09:57 AM

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...

**FernandoRevilla** · November 13th 2010, 10:11 AM

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.

**Educated** · November 13th 2010, 11:54 AM

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

**rtblue** · November 14th 2010, 06:58 AM

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

**Unknown008** · November 14th 2010, 07:26 AM

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

**Wilmer** · November 14th 2010, 08:09 AM

x = ~.56714....

**skeeter** · November 14th 2010, 10:18 AM

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 $x_3$

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