Given two real valued functions $f(x)$ and $g(x)$ which both satisfy the conditions that they are sums of exactly $k$ positive terms each, each term of the form $b_i^x$ where $b_i \in \mathbb{N}$ $(i\in \{1,...,k\})$, can we come up with the maximum number $m$ of points in which $f(x)$ and $g(x)$ intersect (in general)?
Consider $x\in R$, $x\geq 0$ or even $x\in \mathbb{N}_{0}$ if it makes the question simpler.
Examples:
$f(x)=2^x+42^x, g(x)=3^x+4^x$ or $f(x)=30^x+20^x+10^x, g(x)=1^x+2^x+3^x$. In both examples $f$ and $g$ at least intersect at $x=0$ since $f$ and $g$ both have $k$ terms.
My guess was, at first, that it's exactly one point (being $x=0$), but I am not so sure anymore and now consider the answers $m=2$ or some $m(k)\in O(k)$. Is it important to know that the terms are convex to solve this problem?
Edit 2: Since this is very interesting problem to me and I think it fits, can one also find out for what $c$ there cannot be any new intersection for all $x>c$?
Edit: After trying to come up with examples on wolfram alpha that have more than 1 intersection I quickly found multiple examples, but I decided to stop after a while without being able to come up with an example where there are 3 or more intersections.
If you think about this problem as a Laplace transform then I think that there is a simple argument that there are at most $k$ zeros. In all cases, there is at least one zero (for $x=0$) as you point out.
Let's consider the sum $F(s)=\sum_{i=1}^k e^{a_is}-e^{b_is}$. This is actually the Laplace transform of: $$f(t)=\sum_{i=1}^k\delta(t-a_i)-\delta(t-b_i)$$ where $\delta(t)$ is the Dirac delta function. Assuming that all $a_i$ and $b_i$ are positive, we have: $$\mathcal L\{f(t)\}(s)=F(s)$$
So you simply want to find the zeros of $F(s)$.
We can also look at the function $g(t)=\int_0^t f(\tau)d\tau$, which has the Laplace transform $G(s)=\frac1sF(s)$. Thus, both functions $\mathcal L\{f(t)\}(s)$ and $\mathcal L\{g(t)\}(s)$ share the same zeros.
It's much easier to look at $g(t)$ as it's simply a sum of the Heaviside step functions. Then, if you choose the $a_i$ and $b_i$ properly to make alternating boxes (i.e.: going above and below the $t$ axis), then the function $G(s)$ can be made to cross the $s$ axis for each "box". Each box also need to be "wider" than all the preceding ones to ensure that we create a new zero in $G(s)$.
This doesn't provide you with a way to compute an actual example but it shows that you can get at most $k$ zeros.
However, here is an example of a form with three roots: $$e^\frac x{32}-e^\frac x{16}-e^\frac x8+e^\frac x4+e^\frac x2-e^x$$ as you can see here.