It's obvious that $x=0$ is one of the roots. According to the graphs of $e^x$ and $kx + 1$, there's another root $x_1 > 0$ when $k > 1$. Is there a way to represent it numerically.
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I believe there is no closed form in terms of elementary functions. They can be computed numerically, of course. (When $k<1$ there should be second solutions, too, but they should be with $x_1<0$.) – Thomas Andrews Oct 14 '15 at 15:19
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have you tried newton with a start value greater then 1 – Börge Oct 14 '15 at 15:19
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2You are trying to compute the inverse function of $f(x)=\frac{e^x-1}{x}$, essentially. – Thomas Andrews Oct 14 '15 at 15:22
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the solution contains the product logarithmus – Dr. Sonnhard Graubner Oct 14 '15 at 15:26
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If $t = -x - 1/k$, we have $$t e^t = (-x - 1/k) e^{-x} e^{-1/k} = - e^{-1/k}/k $$ The solutions of this are $t = W(-e^{-1/k}/k)$, i.e. $$x = - W(-e^{-1/k}/k) - 1/k$$ where $W$ is one of the branches of the Lambert W function. If $k > 1$, $-e^{-1/k}/k \in (-1/e,0)$, and there are two real branches: the $0$ branch (which gives you $x=0$), and the $-1$ branch, which gives you the solution you want.
Robert Israel
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Thanks a lot, though it's my first time to look into the Lambert W function. – Chielo Newctle Oct 14 '15 at 15:57
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$$e^x=kx+1$$ $$x=\log(1+kx)$$ so $$x=\log(1+k\log(1+k\log(1+k\log(1+....))))$$
E.H.E
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2Assuming that converges, of course. And it might not converge to anything other than $0$. – Thomas Andrews Oct 14 '15 at 15:39
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@ThomasAndrews I have checked it for many values of $x,k$ when they greater than 1, all the cases were converge – E.H.E Oct 14 '15 at 15:45
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1Sorry, but what is your starting value for $x_0$? "I have checked it for many values" is not what we call proof in mathematics, although it is likely true. – Thomas Andrews Oct 14 '15 at 15:46