Solving e^x + x = c Algebraically

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Homework Statement



I am looking for a way to solve the equation e^x + x = c, with c being some constant. I know that I could get an answer from graphing, but I would like to know how to solve an equation like this algebraically. How is this possible?

Homework Equations



None.

The Attempt at a Solution



I have tried manipulating the equation in many ways, but nothing has led to a solution.
 
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Taking ln to isolate the exponent introduces ln(c-x), which requires exponentiation, which requires ln...

I don't think it can be solved algebraically.
 
A (admittedly cheap) approximate solution would be this:

Replace [tex]e^x[/tex] by a truncated series approximation.

[tex] <br /> e^x \approx 1 + x + \frac{x^2} 2[/tex]

You now have

[tex] \begin{align*}<br /> \left(1 + x + \frac{x^2} 2 \right) + x & \approx c \\<br /> \frac{x^2} 2 + 2x + \left(1 - c\right) & \approx 0 \\<br /> x^2 + 4x + 2\left(1 - c \right) & \approx 0<br /> \end{align*}[/tex]

Approximate solutions can be found using the quadratic formula. A graph will help you determine which root to use.
BUT
it may be difficult to determine the accuracy of any solution found in this way. I suspect (but have not investigated numerically) that the closer [tex]c[/tex] is to zero, the closer the approximate solution will be to the true solution.

Short of an approach like this there is (as the others have quite ably stated) no closed form way to solve this problem.
 
"I wonder about what is really wanted by the teacher in terms of solving this."

That's the true rub, isn't it? Of course, I'm not at all sure that the OP was asking a question from a teacher rather than simply one of his/her own investigation. If it came from a teacher or professor I would like to see the original form: I would hope that a teacher would provide a clearer request.
 
statdad said:
... I'm not at all sure that the OP was asking a question from a teacher rather than simply one of his/her own investigation.

Since he has not responded since posing the question 4 days ago, I guess we'll never know for sure.

If the OP shows more interest in getting an answer, I might be willing to spend time explaining how to get an accurate numerical solution.