Maximum number of solutions

Question

While playing in Desmos I noticed something interesting: The following equivalents have a maximum of $n-1$ solutions:

$a_1\cdot b_1^x+a_2\cdot b_2^x+a_3\cdot b_3^x+...+a_n\cdot b_n^x=0$
For example:
$3.14\cdot2^{x}-69\cdot3^{x}+2.71\cdot5^{x}=0$ has $2(=3-1)$ solutions at: $x=-7.619$ and $x=6.330$
How does one prove such a thing?

Answer 1

Assuming that all the $b_j$ are positive and pairwise distinct (else, one could look at equations like $2 \cdot 2^x - 2 \cdot 2^x = 0$) and that all $a_i \neq 0$, consider the function $f \colon \mathbb R \to \mathbb R$ given by

$$f(x) = a_1b_1^x + ... + a_n b_n^x.$$

Now, if $n = 1$, clearly $f$ has no zeros, for any choice of $a_1$ and $b_1$ as above.

Suppose now that we have already proven that

$$\forall a_2',...,a_{n}', \;\; \forall b_2, ..., b_{n} \text{ as above:}\ a_2'b_2^x + ... + a_{n}'b_{n}^x = 0 \text{ has at most } n-2 \text{ solutions}.$$

We want to prove in the inductive step that $f(x) = 0$ has at most $n-1$ solutions.

Since $b_1^x \neq 0$ for all $x$, we can divide the equation $f(x) = 0$ by $b_1^x$ to assume without loss of generality that $b_1 = 1$. The map $f$ is differentiable with

$$f'(x) = a_2 \ln(b_2) b_2^x + ... + a_n \ln(b_n) b_n^x.$$

By induction hypothesis, $f'(x) = 0$ has at most $n-2$ solutions. We can then conclude that $f(x) = 0$ has at most $n-1$ zeros in view of the following question: Proving that if $f′$ has at most $n−1$ zeros, then $f$ has at most $n$ zeros