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I am trying to find a general formula for x and y given that $y=mx+c$ and $y=Ae^{kx}$, with m, c, A and k as constants (and e is Euler's number). essencially, find the point(s) where an exponential function crosses a linear function on a graph. all that I can see can be done is remove y by saying $mx+c=Ae^{kx}$. but then I can see no way of getting x on its own, as it is an exponent and in a polynomial. so a formula for x and y.

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  • $\begingroup$ Just a disclaimer: I've never done much reading in this particular area. I don't think there is an expression for this in terms of elementary functions. I believe what you have is a case of the Lambert W function. $\endgroup$ – JessicaK Mar 26 '15 at 16:09
  • $\begingroup$ Let me WA it for you: link $\endgroup$ – mvw Mar 26 '15 at 16:16
  • $\begingroup$ @JessicaK The WA solution indicates that one important case might need a special function, so you were pointing in the right direction. $\endgroup$ – mvw Mar 26 '15 at 16:19
  • $\begingroup$ thanks for the link! very useful. $\endgroup$ – stanley dodds Mar 26 '15 at 16:43
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As Jessica commented, this kind of equations (which mix polynomial and exponential terms) do not show explicit solutions in terms of elementary functions (this is already the case for $x=e^x$).

However, as Jessica pointed out, the solution can be expressed using Lambert function (I suggest you read the Wikipedia page; to me, it is one of the most important functions after $e^x$ and $\log(x)$ - Lambert and Euler worked together).

In the case you give (assuming that all coefficients are $\neq 0$), the solution write $$x=-\frac{c}{m}-\frac{W(-z)}{k}$$ where $$z=-\frac{A k }{m}e^{-\frac{c k}{m}}$$ In the Wikipedia page, you will find very nice numerical approximations for this functions.

If you cannot use Lambert function, the only way to solve the equation will be to use a root-finding method such as Newton.

To extend a little, any equation which can write $$A+B x+C \log(D+Ex)=0$$ has an explicit solution in terms of Lambert function.

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