Let $a,b \in \mathbb R^n$. I am interested in the following equation:
$$ \sum_{i=1}^{n} a_i e^{b_i x} = 0 $$
where $x\in\mathbb R$, $a = (a_1,\dots,a_n)$ and $b = (b_1,\dots,b_n)$. Edit: $b_i\neq b_j$ for all $i\neq j$.
I am wondering if there is a way to count the number of possible solutions to this equation. I believe the maximum number of solutions is $n-1$. In other words:
$$ \# \{ x \in \mathbb R \mid \sum_{i=1}^{n} a_i e^{b_i x} = 0 \} \overset{?}{\leq} n-1 $$
For $n=2$, the situation is pretty simple. You have the following equation:
$$ a_1 e^{b_1x} + a_2 e^{b_2 x} = 0 $$
Assuming $a_2 \neq 0$, it can be rearranged into
$$ -\frac{a_1}{a_2} = e^{(b_2-b_1)x} $$
Assuming $\frac{a_1}{a_2}$ isn't positive, in which case there are no solutions, we can take logarithm.
$$ (b_2-b_1)x = \log\left(-\frac{a_1}{a_2}\right) $$
And assuming $b_1\neq b_2$, there is a unique solution:
$$ x = \frac{1}{b_2-b_1} \log\left(-\frac{a_1}{a_2}\right) $$
Evidently, depending on the values of $a$ and $b$, the number of solutions is either 0 or 1.
For $n\geq 3$, the above approach doesn't work, but some experimental evidence suggests the limit is 2 zeroes. I tried to convert the equation to a polynomial by making a substitution $y=e^x$, but the result is not a polynomial since the powers on $x$ are arbitrary real numbers. So, how can we prove the number of solutions is bounded above by $n-1$, if it's even true?