Is there a general way to solve equations like the one below? $$ a = b^{y} - c^{y} $$ Where b > 1, c < 1 and a > b? For example...
$$ 5 = 1.1^{y} - 0.1^{y} $$
Moreover, would the problem be any easier to solve if we could choose what b and c were...so long as b>1 and c<1?
If $\frac{\ln b}{\ln c}\in \mathbb{Q}$, then there are $p,q\in\mathbb{Z}$ with $b^p = c^q$ and $q\geq 0$. Let $d:=c^{\frac{1}{p}}=b^{\frac{1}{q}}$. Then $a=b^y-c^y$ becomes $a=d^{yq}-d^{yp}$. If we substitute $t=d^y$, we get $$a=t^q-t^p.$$ If $p$ and $q$ are positive, this is a polynomial of degree $\max\{p,q\}$. If $p$ is negative, we multiply the equation with $t^{-p}$ to get a polynomial of degree $q-p$. We can now try to find the roots of the polynomial and then get the solution by "unrolling" the substitution(s). $$ y=\log_d t = \frac{\ln t}{\ln d} $$ If $\frac{\ln b}{\ln c}\not\in \mathbb{Q}$, then a numerical method must be used.