How can one find the zeroes of $f(x)=ae^{bx}+cx+d$?

Question

A certain physics problem I have been working on has turned into a math problem. Particularly, I want to find the solutions of some equation of the form

$$f(x)=ae^{bx}+cx+d = 0$$

where $a, b, c,$ and $d$ are constant, real numbers that come from the physics problem and $x$ will be a real number. I do not know how to find the solutions of an equation of this form. In the special case where $c=0$, the solution is simply

$$x = \frac{\ln(\frac{-d}{a})}{b}$$

but for other values of $c$, I am stumped. If there is no exact form for the roots, is there a relatively simple way to estimate the roots?

Answer 1

$$ae^{bx}=-(cx+d)$$ Let : $y=-(cx+d)$ hense $x=-\frac{1}{c}(y+d)$ $$ae^{-\frac{b}{c}(y+d)}=y$$ $$ae^{-\frac{bd}{c}}=ye^{\frac{b}{c}y}$$ $$\frac{ab}{c}e^{-\frac{bd}{c}}=\frac{b}{c}ye^{\frac{b}{c}y}$$ Let $X=\frac{b}{c}y$ $$Xe^X=\frac{ab}{c}e^{-\frac{bd}{c}}$$ With the Lambert W function : $$X=W\left(\frac{ab}{c}e^{-\frac{bd}{c}}\right)$$ $x=-\frac{1}{c}(y+d)=-\frac{1}{c}\left(\frac{c}{b}X+d\right)=-\frac{1}{b}X-\frac{d}{c}$ $$x=-\frac{1}{b}W\left(\frac{ab}{c}e^{-\frac{bd}{c}}\right)-\frac{d}{c}$$

Answer 2

By a suitable change of variable, you can turn the equation to

$$e^t=mt+p$$ which describes the intersection of a straight line with the exponential. As described in the post of @jjacquelin, this can be solved using the Lambert function.

If you don't have $W$ handy, you can resort to classical numerical methods.

Deriving the equation,

$$e^t=m$$ gives you the slope of the tangent, wich occurs at $t_0=\ln(m)$. Then, a parallel to the tangent intersects the exponential in $0$ or $2$ points, depending on the value of $p$ compared to

$$p_0=e^{t_0}-mt_0=m-m\ln(m).$$

If $p<p_0$ there is no intersection; if $p=p_0$, a double root at $t=t_0$; otherwise one root $t<t_0$ and another $t>t_0$. You can find initial estimates by solving this quadratic equation for $t$ (derived using Taylor):

$$e^t\approx1+t+\frac{t^2}2=mt+p.$$

The case of $m<0$ is different. There is always a single root and you can start Newton's iterations from any value.

Answer 3

The roots can be computed using any standard root-finding algorithm. For example, with Newton's method, we may use

$$x_{n+1}=x_n-\frac{ae^{bx_n}+cx_n+d}{abe^{bx_n}+c}$$

Without anything really advanced though, we could also use fixed-point iteration:

$$x=\frac1b\ln\left(-\frac{cx+d}a\right)=-\frac{ae^{bx}+d}c$$

one of which will likely converge provided you start sufficiently close to the root you are trying to find.