How can I write $\frac{1}{(a+x)}$ as an exponential function $y = Ce^{-kx}$? How can I write $\frac{1}{a+x}$, $a$ a non-zero positive constant, in exponential terms in the form of $y = Ce^{-kx}$? I've tried to use to Taylor series but they only seem to work for $x < 1$.
 A: So as has been pointed out, this can't be done.
However, let's say that you want to do a good job near $x = 0$. In particular let $f(x) = 1/(a+x)$ where $a$ is fixed in advance; let $g(x) = Ce^{-kx}$ where $C$ and $k$ are what we're trying to find. Then we'll find $C, k$ such that $f(0) = g(0)$ and $f^\prime(0) = g^\prime(0)$. (Why do I find two equations here? Because we have two unknowns.)
$f(0) = g(0)$ and $f^\prime(0) = g^\prime(0)$ are, doing the computation, $1/a = C$ and $-1/a^2 = -Ck$. Plug $1/a = C$ into the second equation and get $k = 1/a$. So the function $g(x) = (1/a)e^{-x/a}$ is, in some sense, a good approximation.
You say you used Taylor series. I don't know what exactly you tried. But one thing you might do in such a context would be to write the Taylor series 
$$1/(a+x) = 1/a - x/a^2 + x^2/a^3 + x^3/a^4 + \cdots$$
and 
$$Ce^{-kx} = C - Ckx + {C k^2 x^2 \over 2} - {C k^3 x^3 \over 6} + \cdots $$
and set the first and second terms of these equal; that of course gives the solution I already gave.
Edited to add: Let's look for approximations of the form $g(x) = \sum_{i=1}^n C_i e^{-k_i x}$ for $n = 2, 3$. For convenience set $a = 1$. When $n = 2$ we have four parameters, so we'll find $C_1, C_2, k_1, k_2$ so that $f(0) = g(0)$ and the first three derivatives are equal at zero. This gives, after some computation (Maple-assisted), approximately
$$g(x) = 0.854 e^{-0.586x} + 0.146 e^{-3.414x}$$
or exactly
$$g(x) = {2 + \sqrt{2} \over 4} e^{-(2-\sqrt{2})x} + {2 - \sqrt{2} \over 4} e^{-(2+\sqrt{2})x}. $$
With $n = 3$ the corresponding approximation (setting the first six derivatives equal) is
$$g(x) = 0.711 e^{-0.416x} + 0.279 e^{-2.294x} + 0.010 e^{-6.290x} $$
Now unfortunately this is not an area I know very well.  I'm actually finding these by this procedure: set $f(0) = g(0)$ and asking Maple to solve for one variable in terms of the other $2n-1$; make this substitution, then set $f^\prime(0) = g^\prime(0)$ and solve for one of the remaining $2n-1$ variables in terms of the remaining $2n-2$; and repeating. Unfortunately the expressions involved get very complicated and Maple is having trouble do the $n = 4$ case symoblically. I'm mentioning this in case anyone recognizes this as some standard procedure in numerical analysis.
A: Stone-Weierstrass theorem is your friend here. 
To wit, the function $f$ defined on $[0,1]$ by $f(0)=0$ and $f(u)=\displaystyle\frac1{a-\log(u)}$ if $u\in(0,1]$ is continuous, hence for every positive $\varepsilon$ there exists a polynomial function $P_\varepsilon$ such that $|f(u)-P_\varepsilon(u)|\le\varepsilon$ for every $u$ in $[0,1]$. Writing $P_\varepsilon$ as
$\displaystyle P_\varepsilon(u)=\sum_{k=0}^{n_\varepsilon}C_k^\varepsilon u^k$ and using this for $u=\mathrm{e}^{-x}$, one gets that, for every nonnegative $x$,
$$
\left|\frac1{a+x}-\sum_{k=0}^{n_\varepsilon}C_k^\varepsilon\mathrm{e}^{-kx}\right|\le\varepsilon.
$$
If one wishes to get rid of the $k=0$ term, one proceeds as follows: the limit when $x\to+\infty$ of the inequality above indicates that $|C_0^\varepsilon|\le\varepsilon$, hence, for every nonnegative $x$,
$$
\left|\frac1{a+x}-\sum_{k=1}^{n_\varepsilon}C_k^\varepsilon\mathrm{e}^{-kx}\right|\le2\varepsilon.
$$
Caveat: nothing says that the coefficients $C_k^\varepsilon$ are nonnegative.
A: You will not find a way to make $\frac{1}{a+x}=C\exp(-kx)$ as they are functions with very different shapes.  They both go to 0 as $x \to \infty$, but $C\exp(-kx)$ does it much faster.
