Determining under what conditions an exponential function is linear I'm working through an exercise to determine when an exponential function of the form:
y = ae^(bx)+c 

is approximately or exactly linear (of the form y=mx+d)
Surely it takes the form y=mx+d when e^(bx) equals x, which solves to b=log(x)/x. But that doesn't lead me anywhere given that b is a constant and log(x)/x is not.
I've also tried to solve the equation: ae^(bx)+c = mx+d, which led me as far as bx-log(x) = (a/m)(d-c+1). I was hoping this would give me an understanding of the problem, but I'm stuck there.
Essentially, I'd like to be able to formulate a set of parameters a, b, and c that lead to the function taking a linear form. Or, to demonstrate mathematically that this is or isn't possible. Thoughts? 
 A: 1. Exponential functions $x\to e^{bx}$ are not linear.
When $a=0$ or $b=0$ the function $x \to a e^{bx} + c$ is constant, which is linear.  To handle the remaining possibilities, assume that $a\ne0$ and $b\ne 0.$  Let's find $(a,b)$ for which this function is linear.  When it is, so is the function $h: x\to e^{bx}.$  However, the latter is not linear on any set of three or more points.
To see why not, let $x_1\lt x_2\lt x_3$ be three distinct points in the domain.  Because $h$ is strictly convex (proof: its second derivative $b^2 e^{bx}$ is strictly positive everywhere), the value of $h$ at the middle point $x_2$ cannot equal the value of the linear function determined by the endpoints $(x_1,h(x_1))$ and $(x_3,h(x_3)),$ whence $h$ is not linear.
Incidentally, all functions defined on sets of just one or two (real) numbers are linear.  The real import of the foregoing is that no function expressible as $x\to e^{bx}$ is linear even on very tiny intervals.  That follows from the foregoing because even tiny intervals contain more than two points.

2. One can approximate exponentials by linear functions.
The interesting aspect of the question concerns how to approximate $y:\,x\to ae^{bx}+c$ by some linear function $x\to mx + d.$   A standard way is through Ordinary Least Squares regression.  Given a probability distribution $F_X$ for $x,$ this method finds numbers $(d,m)$ to minimize the mean squared difference between the values of the function and its approximator; namely,
$$\eqalign{
\text{Mean squared difference}(d,m) &= \int_{-\infty}^{\infty} |y(x) - (mx+d)|^2 \mathrm{d}F_X(x) \\ &= E[(ae^{bX}+c - (mX+d))^2]}\tag{*}$$
where $X$ is any random variable with distribution $F_X.$
As is well known--and constantly referred to on the pages of this site--there is a  solution $(\hat d, \hat m)$ given by the Normal equations and it can be expressed in matrix form as
$$\pmatrix{\hat d \\ \hat m} = \pmatrix{1 & E[X] \\ E[X] & E[X^2]}^{-1} \pmatrix{E[y(X)] \\ E[Xy(X)]}\tag{**}$$
In particular, when it makes sense to change the units of measurement of $X,$ you can standardize it (through a suitable shift and rescaling) to put its mean $E(X)$ at $0$ and establish a unit variance, whence $E(X^2)=1$ and the matrix in $(**)$ becomes the identity; whence, for standardized variables $X,$
$$\pmatrix{\hat d \\ \hat m} = \pmatrix{E[y(X)] \\ E[Xy(X)]}.$$
This reveals $\hat d$ (the intercept) as the mean of the random variable $y(X)$ and $\hat m$ (the slope) as the covariance of $(X,y(X)).$
For example, let $X$ have a standard Normal distribution.  To work out the OLS fit to $ae^{bX}+c,$ compute
$$E[X] = 0,\quad E[X^2] = 1,\quad E[ae^{bX}+c] = ae^{b^2/2}+c,\quad E[X(ae^{bX}+c)]= abe^{b^2/2}$$
giving
$$\pmatrix{\hat d \\ \hat m} = \pmatrix{1 & 0 \\ 0 & 1}^{-1} \pmatrix{ ae^{b^2/2}+c \\ abe^{b^2/2}} = \pmatrix{ ae^{b^2/2}+c \\ abe^{b^2/2}}.$$
Thus, the best linear approximator is the function
$$x \to \hat d + \hat{m} x = ae^{b^2/2}\left(c + bx\right).\tag{***}$$

In this figure, the $x$ coordinates have a Normal distribution with mean $\mu=0$ and standard deviation $\sigma=1.$ The red curve graphs the function $x\to ae^{bx}+c$ with $a=1/2,$ $b=-1/2,$ and $c=1.$  The blue line is the OLS fit $(\text{***}).$
Because a Normal distribution focuses much of its probability between $\mu-\sigma$ and $\mu+\sigma,$ the OLS solution makes the line come relatively close to the curve in this region, as is apparent in the figure.  The OLS objective $(*)$ is the weighted $L^2$ distance between these curves: it heavily penalizes large (vertical) differences compared to small distances and it weights all those differences according to the distribution of $X.$  Thus, it "cares" relatively little about the larger discrepancies more than a few standard deviations away from the mean.

There are, of course, other methods to approximate a function $y$ by a linear function, depending on how you measure the discrepancy between the function and its approximator and depending on what probability distribution you assume for its argument $x.$  The OLS method is particularly simple and tractable, so even when your underlying statistical problem requires a different measure of discrepancy (such as maximum deviation), the OLS solution may serve as a good starting point to search for a solution.
Finally, lest elements of this analysis seem unusual, note that when you have a dataset of the form $(x_1,y(x_1)), (x_2,y(x_2)), \ldots, (x_n, y(x_n))$ for some function $y,$ you may view it as the function $y$ weighted by the empirical distribution of $(x_1,x_2,\ldots, x_n).$ This distribution is discrete, so the expectations in $(*)$ and $(**)$ become summations, which you will recognize as the standard way to express the Normal equations.
A: I think you've basically almost answered the question already. If the function is exactly linear when b=log(x)/x, then it will be approximately linear when b is close enough to log(x)/x. So if you set a cut-off value for how different b is allowed to be from log(x)/x, then that does it.
If b is known then your function will be "approximately" linear for those values of x that bring log(x)/x close enough to b (depending on how you define your cut-off).
