If $a^{2x-1} = b^{1-3y}$ and $a^{3x-1} = b^{2y-2}$, show $13xy = 7x +5y -3.$ If $a^{2x-1} = b^{1-3y}$ and  $a^{3x-1} = b^{2y-2}$, show $13xy = 7x +5y -3.$ 
I apologize in advance if this forum finds this question trivial but I am desperate for any help, will appreciate it. Thanks.

As suggested in J. W. Tanner's comment to the question, assuming that $2x - 1 \neq 0$ and $3x - 1 \neq 0$, then taking appropriate roots of both sides gives that
$$a^{2x-1} = b^{1-3y} \; \Rightarrow a = b^{\frac{1-3y}{2x-1}} \tag{1}\label{eq1}$$
$$a^{3x-1} = b^{2y-2} \; \Rightarrow a = b^{\frac{2y-2}{3x-1}} \tag{2}\label{eq2}$$
Assuming that $b$ is not $-1,0$ or $1$, the powers of $b$ must be equal in \eqref{eq1} and \eqref{eq2}, so
$$\frac{1-3y}{2x-1} = \frac{2y-2}{3x-1} \tag{3}\label{eq3}$$
You can now cross-multiply and simplify to get the requested equality, plus you will also need to show, for the cases not covered by this solution technique, that the equality still holds, or that you can't determine that it does (e.g., if $a = b = 0$ or $a = b = 1$).
