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I'm struggling with solving $x$ for the following equation: it looks in the form as $e^x x^n=1-cx$ where $c$ is just some constant number.

I tried to solve $x$ explicitly. But it seems not be able to do so. Do I have to solve it analytically using some Newton-Raphson method?

Many thanks!

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    Newton-Raphson method is a numerical method and you have to specify a value for c.2012-08-31

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Consider $f(x) = e^x x^n - (1-cx)$ where $n \ge 1$. $f(0) = -1$. $f(1) = e - 1 +c $, so if $c > 1-e$ there is a root between $0$ and $1$.

If $ c \le 1-e$, $f(x) = e^x x^n - 1 -|c|x$, so, since $e^x > 1+x$ for $x > 0$, if $x_0 = \max(1, |c|^{1/(n-1)})$, $x_0^{n-1}\ge |c|$ so $x_0^n \ge c x_0$ and $f(x_0) > 0$ so there is a root between $0$ and $x_0$.

In either case, we know an interval where there is a positive root.

If $c > 0$, then $f'(x) = e^x x^{n-1}(x+n) + c > 0$, so $f$ is increasing and there can be only one positive root and almost any root finder will do.

If $c \le 0$, we might have to do more, but it is late and I am tired.

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I don't think there's any straight-forward technique to analytically solve for $x$ in those equations, but there are many ways you can ask for fixed-point methods. Just try to put $x = f(x)$ (there are infinitely many ways to do that...) for some function $x$ and do the math. If you don't know how I suggest an introductory course in numerical analysis, which I don't plan on doing here for obvious reasons. =)

Example : $x = \frac{1 - e^x x^n}c$, or $x = \frac{1-cx}{e^x x^{n-1}}$, or $x = e^x x^n - 1 + (c+1)x$.

If this is just personal then I guess you can always ask Mathematica numerical approximations, but in general I don't expect any pattern in the solution as a function of $n$...

Hope that helps,