Number of solutions to an exponential equation

Question

I am trying to prove that there are exactly $0$ or $1$ solutions in $x$ to the following equation:

$A_0 + A_1B^{x} + A_2B^{2x} + ... + A_kB^{kx} = 0$

All elements are real numbers, and $k$ is a positive integer.

It seems true, but I am getting nowhere proving it.

Actually, monotonacity of the left side of the equation would be an ideal result for what I am doing, so if the hypothesis could be strengthened to that, it would be great.

NOTE: I did not translate this hypothesis from its original context correctly. I have updated it. I understand that this totally changes the question, I apologize for the mis statement.

Answer 1

Set $y = B^x$. Upon substitution, the left-hand side is a degree $k$ polynomial in $y$. You should expect $k$ solutions.

If you constrain $B$ or $x$, you may be able to reject some of the roots of the polynomial as solutions to your equation. However, it is not generally hard to arrange, for what I suspect is the most likely constraint you might have had in mind, for all $k$ roots of the polynomial to be distinct positive real numbers, or even distinct positive integers.