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

In summary, the person is trying to solve an equation for x in which x+e^x=4. They use a graphing calculator and find that x=1.0737... approximately. They then try different values of x and narrow in on a value. They use a Newtons approximation formula or a graphing software to get the result. They use the Lambert W function to get y=W(e^4). Finally, they apply the inverse function of f to both sides of the equation to get x=4-W(e^4).
  • #1
allamid06
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Homework Statement


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

Homework Equations

The Attempt at a Solution

 
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  • #2
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.
 
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  • #3
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].
 
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  • #4
HallsofIvy said:
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].

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-
 

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

What is the meaning of "I'm a little rusty, cant solve x+e^x=b"?

This phrase typically refers to a person's lack of practice or proficiency in solving mathematical equations, specifically in this case, an equation involving exponential functions.

What is x+e^x=b and how do I solve it?

x+e^x=b is a mathematical equation known as an exponential equation. To solve it, you can use algebraic techniques such as isolating the variable on one side of the equation and using logarithms to simplify the exponential term.

Why is it difficult to solve x+e^x=b?

This equation can be difficult to solve because it involves both a variable and an exponential function, which can be challenging to manipulate algebraically. It may also require knowledge of logarithms and their properties.

What are some strategies for solving x+e^x=b?

One strategy for solving this equation is to isolate the variable on one side of the equation and use logarithms to simplify the exponential term. Another strategy is to graph both sides of the equation and find the point of intersection, which represents the solution.

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Being able to solve equations like x+e^x=b is important for many fields of study, including science, engineering, and finance. It allows us to model and understand relationships between variables in various systems and make predictions based on those relationships.

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