Prove/disprove that $f(x)=\sum_{i=1}^n c_i e^{-d_i x}$ has at most three zeros

Question

Given the function $$ f(x)=\sum_{i=1}^n c_i e^{-d_i x} $$ where $n\ge 3$ and $d_i\ge 0$, prove or disprove that $f(x)$ has at most three zeros, that is $f(x)=0$ on at most three points.

This looks tedious to answer (no counter-example found). Any hint on how to proceed?

Answer 1

Note that $$p(x)=(x-1)(x-2)(x-3)=x^3-6x^2+11x-6$$ has three positive zeroes $1,2,3$. Therefore, $$ e^{-3x}-6e^{-2x}+11e^{-x}-6e^{-0x}$$ has zeroes $-\ln1,-\ln 2,-\ln 3$.