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I'm a little rusty, cant solve x+e^x=b

  1. Mar 7, 2015 #1
    1. The problem statement, all variables and given/known data
    Hi, I'm new here. I'm really rusty, I resume my career this year, and I'm reading 'the spivak book', (for Calculus 1).
    Making some exercises, I get curious about how to solve this: x+e^x=4
    I would love if someone could give me any trick

    2. Relevant equations


    3. The attempt at a solution
     
  2. jcsd
  3. Mar 8, 2015 #2

    jedishrfu

    Staff: Mentor

    Welcome to PF!

    I don't think there is an analytical way to solve for x in this example. I did graph it and found that x=1.0737... approximately.

    You could find this by trying different values of x like 0,1,2... And then narrowing in the values. You could also uses a newtons approximation formula or a graphing calculator to get the result.

    Try doing it on a graphics calculator or using some graphing software like freemat on a PC or Mac.
     
  4. Mar 8, 2015 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    For an "analytical solution", you can first write the equation as [itex]e^x= 4- x[/itex], then divide both sides by [itex]e^x[/itex] to get [itex](4- x)e^{-x}= 1[/itex]. Now, let y= 4- x so that -x= y- 4. In terms of y, the equation is [itex]ye^{y- 4}= ye^y(e^{-4})= 1[/itex]. Multiply both sides of the equation by [itex]e^4[/itex] to get [itex]ye^y= e^4[/itex].

    Finally, apply the "Lambert W function", which is defined as "the inverse function to [itex]f(x)= xe^x[/itex]", to both sides of the equation getting [itex] y= W(e^4)[/itex]. Since x= 4- y, the solution to the original equation is [itex]x= 4- W(e^4)[/itex].
     
    Last edited: Mar 8, 2015
  5. Mar 8, 2015 #4
    Thanks Hallsoflvy, and thanks jedishrfu. First of all, I will get the Hallsoflvy answer, because, I'm pretending to use anything but the first cap, of the Spivak.
    By the other part, Hallsoflvy, I did that, but then I realize, that Lambert W function was too over the first cap of the book. Anyway I really like the way you explain me that method.
    To be honest, the original problem was [itex]x+3^{x}<4[/itex]. I use desmos.com a lot, to graph these things. I could solve this by seeing that [itex]3^{x}=3[/itex] only if [itex]x=1[/itex], and since [itex]3^{x}[/itex] rises, (by "common sense"), [itex]3^{x}<3[/itex] when [itex]x>1[/itex] and at the same time I can add these and get [itex]x+3^{x}<1+3[/itex]. I could apply the same to show [itex]x+3^{x}>4[/itex] when [itex]1>x[/itex].
    But when I was doing this, I just get curious about how would I find the root, in this case. And then I went (don't know why) to [itex]x+e^{x}=b[/itex] thinking it would be easier.
    PD:Thanks for your time. Since I study software engineer, I will be more interested in Calculus, Discrete Maths,
    Probability and Statistics etc. But also I will have Pysics I. What I'm saying is that I hope to be useful in this community, and I hope, there where place to my questions.
    Also I'm from Uruguay, my apologies for my english. -I can understand you but I'm not really good writing-
     
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