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Math Help - Solving xe^x = 1

  1. #1
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    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!
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  2. #2
    Member rtblue's Avatar
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    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
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  3. #3
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    Quote Originally Posted by rtblue View Post
    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...
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Yehia View Post
    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.
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  5. #5
    Senior Member Educated's Avatar
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    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
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  6. #6
    Member rtblue's Avatar
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    Look at the given function. If x = 0, then the whole left side becomes one. Anything to power zero is one.
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  7. #7
    MHF Contributor Unknown008's Avatar
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    Yes, but if x becomes 0, you have 0e^0 = 0
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  8. #8
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    x = ~.56714....
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  9. #9
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    Quote Originally Posted by Yehia View Post
    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
    Last edited by skeeter; November 14th 2010 at 04:24 PM. Reason: fixed sign typos
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