Existence and Uniqueness of the exponential interpolation Suppose function has the form $$f(x) = ae^{bx}+ce^{dx}$$
where $a,b,c,d $ are real variables. Given four real numbers $x_1,x_2,x_3,x_4$, and four corresponding values $y_1,y_2,y_3,y_4$. 
Whether can we find $a,b,c,d$ such that $$f(x_j)=y_j,j=1,2,3,4$$
 A: No, this is in general not possible.
Here is a simple counter example:
Consider a case where two of the chosen points would be $(1,0)$ and $(1,0)$.
This gives:
$$
f(-1) = a e^b + c e^d = 0
$$
$$
f(1) = a e^{-b} + c e^{-d} = 0
$$
Obviously the function $f(x)=0$ would be valid. For a non-trivial solution, one of the values $a$ or $c$ needs to be non-zero. Without loss of generality, we examine the case $a\neq 0$. If we rewrite the two equations, we get 
$$
-c/a = e^{b-d} = e^{d-b}
$$
from which it follows that $d=b$ and $c=-a$. Consequently, we find that
$$
f(x)=a e^{bx} - a e^{-bx} = 2 a \sinh bx
$$
With $f(1)=0$ this gives $b=0$ and we find that again $f(x)=0$. Hence any other additional point $(x_3,y_3)$ with $y_3 \neq 0$ would make it impossible to find a solution for $a,b,c,d$.
Hence, in general there is no solution.
A more general way of looking at the question is to consider the derivative of the function $f(x)$:
$$
f'(x) = a b e^{b x} + c d e^{d x}
$$
If we look for the solutions $f'(x)=0$, we find that there is at most a single solution for $x$. (Ignoring the trivial function $f(x)=0$). Hence, any four points $(x_i,y_i)$ with $x_1<x_2<x_3<x_4$ and where the corresponding $y$ alternatingly increase and decrease, could never be fitted by a function of this particular form. 
