Solving transcendental equation involving exponential functions I'm trying to numerically solve the following transcendental equation:
$$(\alpha x + \beta)=\delta e^{\gamma x}$$
with $\alpha$, $\beta$, $\gamma$ and $\delta$ real, positive constants. It is equivalent to finding the intersection between the line $y=(\alpha x + \beta)$ and the exponential function $y=\delta e^{\gamma x}$. For a proper choice of the constants, this equation should have two solutions. Graphically, one can verify that, in case it has solutions, the solutions could be found by a "successive approximation" process. However,
$$ x = \frac{\delta}{\alpha}e^{\gamma x}-\frac{\beta}{\alpha}$$
does not seems to me to be a contraction. How could I show that a "successive approximation" process could be convergent for the above?
Thank you very much in advance.
 A: The only general procedure that is a contraction sufficiently close to the solution is the Newton's method. The method gives the contraction:
$$x-\frac{\delta\,{e}^{x\,\gamma}-\alpha\,x-\beta}{\delta\,\gamma\,{e}^{x\,\gamma}-\alpha}$$
To derive the above formula, let
$$ f(x) = \alpha x + \beta - \delta e^{\gamma x}$$
and you are looking for $f=0$.
Newton's method gives 
$$
x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
$$
Newton makes no guarantees though. If the problem is well posed in the sense that the solution is not a critical point (i.e. $f'=0$ at the solution) then one can be reasonably sure that the method will give the solution if you start with a good initial guess. Refer to any standard text on numerical analysis for more.
A: From my numerical analisys text:
If $g(x)$ is continous on $[a,b]$ an $g(x) \in [a,b] \ \  \forall x \in  [a,b]$ then $g$ has a fixed point, i.e. $g(x)= x$ for some $x \in [a,b]$. If furthermore $|g'(x)|<1 \  \forall x\in (a,b)$ then the fixed point is unique.
you already have that $g(x) = \frac{\delta}{\alpha}e^{\gamma x}-\frac{\beta}{\alpha}$ is continus, so you just have to show that $g(x) \in [a,b]$
