What kind of function space does the a set of linearly independent exponential form? I'm confused because you can define a basis $\{1, x, x^2,\ldots\}$ as the basis for the space of polynomials, $\{1, \sin(x), \cos(x),\ldots\}$ as the basis for fourier series, but little words are said for other basis i.e. $\{\exp(x), \exp(2x), \exp(3x),\ldots\}$.
Could you say that any linearly independent set of function defines a basis on some of space or is this a conflation of terminology in linear algebra and functional analysis?
 A: The set $\{1,x,x^2...\}$ forms an algebraic basis of the vector space of all polynomials. (Every polynomial can be written as a finite linear combination of these elements)
However, the Fourier basis $\{1\}\cup\{sin(nx),cos(nx), n \in \mathbb{N} \}$ is not an algebraic basis for the vector space $L^2[0,2\pi]$. Rather, since it is a complete orthonormal set, we can express every vector $f$ in terms of its fourier series
$$ f = \sum_{n \in \mathbb{N}} f_n sin(nx) + \sum_{n \in \mathbb{N}} g_n cos(nx)$$
Note that the convergence here is not pointwise: it is convergence in the $L^2$ norm.
So your question about what kind of function space is generated by the set of exponentials depends on whether you are looking at an algebraic basis (a Hamel basis), or in the analytic sense : take the algebraic span of the set and look at its closure.
BUT, since we are talking about the "closure", you should have some topological structure underlying. For instance, if you consider $\{exp(nx),n\in \mathbb{N}\}$ as a subset of $L^p[0,1]$, and look at the span closure, what you end up with will depend on the $p$ you started with.
A: Note that $\{\mathrm{e}^{0x}, \mathrm{e}^{1x}, \mathrm{e}^{2x}, \mathrm{e}^{3x}, \dots \} = \{1, u, u^2, u^3, \dots \}$ where $u = \mathrm{e}^x$.  Unless the everywhere-positivity of $\mathrm{e}^x$ on $\Bbb{R}$ captures something essential about the space of functions you're interested in, there's nothing new here.
Also, $\mathrm{e}^{kx} = \cos (\frac{k}{\mathrm{i}} x) + \mathrm{i}\sin (\frac{k}{\mathrm{i}} x)$, so again this may not capture something you don't already have.
A: The answer begs the question--but hopefully this will clear up the analogous terminology a bit. The definition of the span of a set of vectors (or functions) is the intersection of the subspaces generated by the linear combination of those vectors. 
A basis for a given space is thus defined as a set of vectors or functions whose span is the space itself. An efficient basis, then, is effectively a set of orthonormal vectors or functions in that group. In functional space, analogous to the dot product for vectors, two functions are orthogonal iff their inner product (the integral of the product one function and the complex conjugate of the other) is zero (naturally these are linearly independent). 
Thus an efficient basis for a space is a set of orthonormal functions whose span is the space itself. 
So, no, not just any set will do as a functional basis. The set must minimally span the space. An added bonus would be not wasting room with vectors that are not orthonormal--but in practice this is sometimes unavoidable (measurement is commonly a dirty practice). 
