Let $a\_1, a\_2, \dots, a\_n$ and $b\_1, b\_2, \dots, b\_n$ be real numbers, and assume that $\{a\_i\} \neq \{b\_i\}$. Can the equation
$$ e^{a\_1 x} + e^{a\_2 x} + \dots + e^{a\_n x} = e^{b\_1 x} + e^{b\_2 x} + \dots + e^{b\_n x}$$
have more than $n$ real roots including $0$ and counting multiplicities? 


There are special cases that might be familiar to some. It cannot have $(n+1)$ roots at $0$ because that would imply $\sum a\_i^k = \sum b\_i^k$ for all $0 \le k \le n$. Likewise, the equation cannot have all of $0, r, 2r, \dots, nr$ as roots. In fact, if $0, r, \dots, (n-1)r$ are roots, then it can be proved that there are no additional roots. 



Number of real roots of an exponential polynomial
Seems that it can not have more than $n$ roots. Let us use the following generalization of Descartes rule for signed measures. Namely, let $\mu$ be a Borel signed measure on a real line, with compact support (this condition may be of course weakened). We say that $\mu$ has at most $k$ changes of sign if there exist points $c\_1<c\_2<\dots <c\_k$ such that restriction of $\mu$ to each of the intervals $(-\infty,c\_1],(c\_1,c\_2],(c\_2,c\_3],\dots,(c\_k,\infty)$ is either non-negative or non-positive. I claim that in this situation Laplace transform $L(t):=\int e^{tx} d\mu(x)$ has at most $k$ real roots with multiplicity counted (except the case $\mu\equiv 0$). Induction in $k$. Base $k=0$ is clear. Assume that $k\geq 1$, for $k-1$ it is established, we should prove for a measure $\mu$ having exactly $k$ sign changes. If $L(t)$ has $N$ roots, $g(t):=e^{c\_1t}(L(t)e^{-c\_1t})'$ has $N-1$ roots, but $g(t)=\int (x-c\_1)e^{tx} d\mu(x)$, and new signed measure $d\nu(x)=(x-c\_1)d\mu(x)$ has at most $k-1$ sign changes (if new measure $\nu$ is 0, then $\mu$ was proportional to the delta-measure in $c\_1$, and $L(t)$ does not have roots at all.) So, by induction hypothesis $N-1\leq k-1$, as we need. 


Now we say $e^{at}-e^{bt}={\rm sign}\,(a-b)\cdot t\cdot \int\_{[a,b]} e^{tx} dx$, so if we divide our equation $\sum\_{i=1}^n e^{a\_it}-e^{b\_it}=0$ by $t$, it becomes a Laplace transform for a measure with density $\sum \pm \chi\_{[a\_i,b\_i]}$, which has at most $n-1$ sign changes. Thus our equation has at most $n-1$ non-zero roots as desired.


(PREVIOUS VERSION, working for simple roots) It is equivalent to counting positive roots of $\sum x^{a\_i}=\sum x^{b\_i}$. There is always root 1, so subtract $n$ from both parts and divide by $x-1$. I hope that after this the difference has at most $n-1$ sign changes and so Descartes rule is applicable (if we assume that exponents are positive integers, as we may assume: having more than $n$ roots is open condition, so exponents may be considered rational, then by scaling change of variables integer.) Indeed, sign change appears if there were more $a$'s than $b$'s on some initial segment, but it became more $b$'s than $a$'s. It requires two points at least. 


UPD: when I say that having more than $n$ roots is an open condition, I, strictly speaking, lie. It is so for simple roots, but for multiple roots it is not: say, small perturbations of $x^2$ may have no real roots. However I believe that the statement is true for multiple roots too, but it have to be fixed somehow. 


