I'm looking for the $x$ such that
$$ -\frac{1}{q^x}+1=a^x -1$$
For any $q$ and $a$
A.K.A the $x$ at the intersection of the two functions. Am I asking the wrong question here?
One obvious solution is $x=0$ but there is another solution I can't seem to find in terms of $q$ and $a$
$0$ is the only solution that works for all values of $q$ and $a$. Let $f(x) = - \frac{1}{q^x}+1$ and $g(x) = a^x -1$, and take $q=1$ and $a=2$ as a simple example. Then $f(x) = 0$ for all $x$, hence $h(x) = f(x) - g(x) = 1-2^x$. Note that for positive $x$, we have $2^x > 1$, so clearly $h(x)$ is always negative for positive $x$, then crosses the $y$-axis at zero. For negative $x$ it's equivalent to $1 - \frac{1}{2^x}$, which is always positive. No other zeroes apart from $h(0)$ are possible.