It is easy to determine a coefficient 'c' of exp(c*x), just log it and find slope. Or if it's
exp(c1*x) + exp(c2*x)
then after log from 0 to the right of left we would find 'c1' and 'c2'. But what if we have more terms? For example, such a sum
exp(x) + exp(1.5*x) + exp(2.5*x) + exp(3*x)
if you loged it, almost would have no difference between
exp(x) + exp(3*x)
So, please, how could I find them in another way?
Best Answer
If you know, that $$y(x) = \exp(c_1 x)+\exp(c_2 x)+\cdots+\exp(c_n x),$$ then calculate $y(1), y(2), \ldots, y(n)\;$: $\;p_1 = y(1), p_2 = y(2), \ldots, p_n=y(n)$.
You'll get system of equations:
$$\left\{ \begin{array}{r} \exp(c_1)+\exp(c_2)+\cdots+\exp(c_n)=p_1; \\ \exp(2c_1)+\exp(2c_2)+\cdots+\exp(2c_n)=p_2; \\ \cdots \cdots \cdots \qquad \qquad \qquad \qquad \\ \exp(nc_1)+\exp(nc_2)+\cdots+\exp(nc_n)=p_n. \end{array} \right. $$
If denote $s_1=\exp(c_1), s_2=\exp(c_2), \ldots, s_n=\exp(c_n)$, then $$\left\{ \begin{array}{r} s_1+s_2+\cdots+s_n=p_1; \\ s_1^2+s_2^2+\cdots+s_n^2=p_2; \\ \cdots \cdots \cdots \qquad \qquad \\ s_1^n+s_2^n+\cdots+s_n^n=p_n. \end{array} \right. $$
With the help of Power sum symmetric polynomial you'll find values $e_1,e_2, \ldots, e_n$ $-$ elementary symmetric polynomials:
$$\left\{ \begin{array}{l} e_1 = s_1+s_2+\cdots+s_n = p_1; \\ e_2 = s_1s_2+s_1s_3+\cdots+s_{n-1}s_n = \dfrac{1}{2}(p_1^2-p_2); \\ \cdots \cdots \cdots\qquad \qquad \\ e_n = s_1s_2\cdots s_n=\dfrac{1}{n}\sum\limits_{j=1}^n (-1)^{j-1}e_{n-j}p_j; \end{array} \right. $$
so $s_1,s_2,\ldots,s_n$ are roots of equation
$$s^n-e_1\cdot s^{n-1}+\cdots+(-1)^{n-1}e_{n-1}\cdot s+(-1)^n e_n=0.$$
Finally, when you will find roots $s_1,s_2,\ldots,s_n$,
$$c_j=\ln(s_j), \; j=1,\ldots,n.$$