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anonymous

  • 5 years ago

how do you solve e^x = x

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  1. anonymous
    • 5 years ago
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    by taking ln for both sides

  2. anonymous
    • 5 years ago
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    Use Newton-raphson method

  3. anonymous
    • 5 years ago
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    what is the newton-raphson method?

  4. anonymous
    • 5 years ago
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    taking the ln does not anything

  5. anonymous
    • 5 years ago
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    please help nowhereman

  6. nowhereman
    • 5 years ago
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    there is no analytic solution, so you should use an approximation method like Newton's tangent method

  7. anonymous
    • 5 years ago
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    how do you know there is no analytic solution?

  8. anonymous
    • 5 years ago
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    you're right. ln does not solve the problem

  9. nowhereman
    • 5 years ago
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    oh wait, there is no solution at all, because \[ e^x >= 1 + x \]

  10. anonymous
    • 5 years ago
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    ya, e^x always larger than x, so e^x = x does not have any solution

  11. anonymous
    • 5 years ago
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    you are wrong. The answer is nonreal, which i am well aware of. I know what the number is to many decimal places. (Just take the ln of any number over and over again), but i have no idea how to find the number.

  12. anonymous
    • 5 years ago
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    you have to iterate using computer methods to find this value

  13. anonymous
    • 5 years ago
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    can you prove that?

  14. anonymous
    • 5 years ago
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    If you use the moivre's theorem you know that e^(ix) = isinx + cosx or something like that

  15. anonymous
    • 5 years ago
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    I used software program and found that x=-W(-1), where W(z) is the product log function

  16. anonymous
    • 5 years ago
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    what software did you use? did u use wolfram alpha? what is the product log function?

  17. anonymous
    • 5 years ago
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    yes.. here is the definition of the function http://www.wolframalpha.com/input/?i=product+log+function

  18. anonymous
    • 5 years ago
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    ok

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