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

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1. Mar 7, 2015

### allamid06

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. Mar 8, 2015

### 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.

3. Mar 8, 2015

### HallsofIvy

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

Finally, apply the "Lambert W function", which is defined as "the inverse function to $f(x)= xe^x$", to both sides of the equation getting $y= W(e^4)$. Since x= 4- y, the solution to the original equation is $x= 4- W(e^4)$.

Last edited by a moderator: Mar 8, 2015
4. Mar 8, 2015

### allamid06

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 $x+3^{x}<4$. I use desmos.com a lot, to graph these things. I could solve this by seeing that $3^{x}=3$ only if $x=1$, and since $3^{x}$ rises, (by "common sense"), $3^{x}<3$ when $x>1$ and at the same time I can add these and get $x+3^{x}<1+3$. I could apply the same to show $x+3^{x}>4$ when $1>x$.
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 $x+e^{x}=b$ 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-