Always transcendental? $a + b e^{-x} - f(x) = 0$ I want to choose $f(x)$ such that the equation $$a + b e^{-x} - f(x) = 0$$ is analytically solvable. Ideally, I want $f(x)$ to be some function that is symmetric about 0 and everywhere positive, like $x^2$ or $|x|$. Is there a function with these properties that renders the equation solvable? 
 A: $$f(x)=\cosh(x)$$satisfies your requirements.
A: There is a closed form solution to
$$
a + \exp(-x)= cx^n
$$
with $n=2$ in terms of generalized Lambert functions.
This is no great surprise as Maple or Mathematica solves the problem with $n=1$:
Solve[a + Exp[-x] == c*x, x]

{{x -> (a + c*ProductLog[1/(c*E^(a/c))])/c}}

ProductLog being LambertW.
Now the $n=2$ case can be rewritten
$$
(x-s)(x+s)\exp(x)=\frac{1}{c} \qquad \text{with } s=\sqrt{\frac{a}{c}},
$$
and this is of the form treated in
I. Mező and A. Baricz,
On the generalization of the Lambert $W$ function with applications in theoretical physics.
arXiv:1408.3999v1.pdf (18 Aug 2014).
A: I'm not sure if this is in the spirit of the question but $f(x) = 1$
Also you can make a function symmetric by defining it piecewise: e.g. $f(x) = \text{if $x > 0$ then }e^{-x} \text{ else } e^x$.
A: If $f(x)$
is  a  polynomial
of degree at most four
in $e^{-x}$,
then the equation
can be solved
by the substitution
$y = e^{-x}$
and then
solving the equation.
This can also be done
for
$f(x)$ being
various combinations of
$e^x$ and $e^{-x}$
such as
$\sinh(x)$,
$\cosh(x)$,
and
$\tanh(h)$.
If the resulting equation
is quadratic in
$e^x$ or $e^{-x}$,
the solution is easier.
If $f(x) = e^{2x}$,
then the equation becomes
a cubic in $e^x$,
which is harder.
