The general description of all continuous functions which are uniform limits of fewnomials seems to be given by the following

**Theorem.** Fix $0<a<b$. The following two properties of a continuous function $f\in C[a,b]$ are equivalent:

(i) $f$ is log-quasipolynomial, that is, $f(x)=P(x,\log x)$ for a certain quasi-polynomial $P(u,v)=\sum_{i=1}^N c_i u^{\kappa_i} v^{m_i}$, where $\kappa_i$'s are real and $m_i$'s non-negative integers;

(ii) $f$ is a uniform limit of a sequence of fewnomials (that is, of functions of the form $f_n(x)=\sum_{k=1}^K \beta_k^{(n)} x^{\alpha_k^{(n)}}$ with the same $K$).

**Proof.** $(i)\Rightarrow (ii)$ follows from $\log x=\lim n(x^{1/n}-1)$ uniformly on $[a,b]$.

$(i)\Rightarrow (ii)$ Induction in $K$. Base $K=0$, then $f_n(x)\equiv 0$ and $f\equiv 0$ too.

Induction step from $K-1$ to $K$. Passing to a subsequence, we may suppose that $\alpha_K^{(n)}$ has a limit $\alpha_K$ which may be equal to $+\infty$, $-\infty$ or be a real number.

If, say, $\alpha_K=\infty$, we may fix $c\in (a/b,1)$ and consider the functions $g_n(x)=f_n(cx)-c^{\alpha_K^{(n)}}f_n(x)$ (it has at most $K-1$ monomials), they converge to $f(cx)$ uniformly on $[a/c,b]$. Thus by induction hypothesis $f(cx)$ is log-quasipolynomial on $[a/c,b]$. In other words, $f$ is log-quasipolynomial on $[a,cb]$. Such log-quasipolynomials for distinct $c$ must be consistent, so $f$ is log-quasipolynomial on $[a,b]$. Analogously if $\alpha_K=-\infty$.

So assume that $\alpha_K$ is a real number. Consider the functions $\tilde{f}_n(x)=x^{-\alpha_K^{(n)}}f_n(x)=\sum_{k=1}^K \beta_k^{(n)} x^{\tilde{\alpha}_k^{(n)}}$, where $\tilde{\alpha}_k^{(n)}=\alpha_k^{(n)}-\alpha_K^{(n)}$. They uniformly on $[a,b]$ converge to $\tilde{f}(x):=x^{-\alpha_K}f(x)$. If $\tilde{f}(x)$ is log-quasipolynomial, then so is $f$. Changing notations back (omit tilde), we now on suppose that $\alpha_K^{(n)}=0$.

Again, we fix $c\in (a/b,1)$ and consider the functions $g_n(x)=f_n(cx)-f_n(x)$. They converge to $f(cx)-f(x)$ uniformly on $[a/c,b]$. By induction hypothesis $f(cx)-f(x)$ is a log-quasipolynomial on $[a/c,b]$. Any log-quasipolynomial may be represented as $H(cx)-H(x)$ for another log-quasipolynomial $H$. So we may write $f(cx)-f(x)=H_c(cx)-H_c(x)$ for a log-quasipolynomial $H_c$. We may assume additionally $H_c(b)=f(b)$. Note that $$H_c(cx)-H_c(x)=f(cx)-f(x)=\sum_{j=0}^{m-1} f(c^{(j+1)/m}x)-f(c^{j/m}x)=H_{c^{1/m}}(cx)-H_{c^{1/m}}(x),$$
thus $p(x):=H_c(x)-H_{c^{1/m}}(x)$ satisfies $p(b)=0$ and $p(cx)=p(x)$ for $x\in [a/c,b]$. This yields $H_c\equiv H_{c^{1/m}}$. If $c_1,c_2$ are rational powers of 2, then there exist positive integers $m_1,m_2$ such that $c_1^{1/m_1}=c_2^{1/m_2}$. Thus $H_{c}=:H$ does not depend on $c$ on a dense set $D\subset [a/b,1]$ of values of $c$'s. So the function $f(x)-H(x)=:F(x)$ satisfies $F(b)=0$, $F(x)=F(cx)$ for $c\in D$. Since $F$ is continuous, this yields $F\equiv 0$ and $f=H$. $\square$

Of course $\exp(x)$ is not log-quasipolynomial on an interval (for instance, because it grows faster at infinity, and if two functions which are analytic on the right half-plane coincide on an interval, they do coincide on the whole half-plane).