Solving for $n$ when in exponents of differences

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I have an equation where

$$ \frac{(c^n + y)}{(a^n - b^n)} = \frac{C}{(a-b)} $$

In my case, $ y << c^n $ , so I could potentially simplify the equation to:

$$ \frac{c^n}{(a^n - b^n)} = \frac{C}{(a-b)} $$

or further

$$ \left(\frac{a}{c}\right)^n - \left(\frac{b}{c}\right)^n = \frac{(a-b)}{C} $$

I have everything except $n$.

Can I solve for $n$, given $a$, $b$, $c$ and $C$?

Can I solve for $n$, given $a$, $b$, $c$, $y$ and $C$ without having to resort to the approximation and taking away $y$?