Show that the equation $a_1e^{\alpha_1x} + a_2e^{\alpha_2x} + \cdots + a_ne^{\alpha_nx} = 0$ has at most $n – 1$ real roots.

For non-zero $a_1, a_2, \ldots , a_n$ and for $\alpha_1, \alpha_2, \ldots , \alpha_n$ such that $\alpha_i \neq \alpha_j$ for $i \neq j$, show that the equation
$$a_1e^{\alpha_1x} + a_2e^{\alpha_2x} + \cdots + a_ne^{\alpha_nx} = 0$$
has at most $n – 1$ real roots.

I thought of applying Rolle’s theorem to the function $f(x) = a_1e^{\alpha_1x} + a_2e^{\alpha_2x} + \cdots + a_ne^{\alpha_nx}$ and reach a contradiction, but I can’t find a way to use it.

Solutions Collecting From Web of "Show that the equation $a_1e^{\alpha_1x} + a_2e^{\alpha_2x} + \cdots + a_ne^{\alpha_nx} = 0$ has at most $n – 1$ real roots."

Hints: Between any two roots of $f(x)$, there lies a root of $f'(x)$ by Rolle’s theorem.

Now, we shall use induction and the above fact to do your question.

For $n=1$- no real root.

Let’s consider $n=2$ in detail for the idea of the general case.

$f(x)=a_1e^{\alpha_1x}+a_2e^{\alpha_2x}$.

Let $g(x)=e^{-\alpha_1x}f(x)=a_1+a_2e^{(\alpha_1-\alpha_2)x}$. What’s $g'(x)?$ How many real roots does $g'(x)$ have? (answer: $0$ real roots). So, $g(x)$ can have at most one real root- which means $f$ can have at most one real root.

Now, use induction for the general case after considering $g(x)=a_1e^{-\alpha_1x}f(x)$. (Note that the real roots of $f$ and $g$ coincide.)