Solve the following system of equations: $\begin{cases} y - \log_3 x = 1 \\ x^y = 3^{12} \end{cases}$
Given system of equations:
$y - \log_3 x = 1$ $\;\;\; \cdots \; (1)$
$x^y = 3^{12}$ $\;\;\; \cdots \; (2)$
Equation $(1)$ $\implies$ $\log_3 x = y - 1$
i.e. $\;$ $x = 3^{y - 1}$ $\;\;\; \cdots \; (3)$
In view of equation $(3)$, equation $(2)$ becomes
$\left(3^{y - 1}\right)^y = 3^{12}$
i.e. $\;$ $y \left(y - 1\right) = 12$
i.e. $\;$ $y^2 - y - 12 = 0$
i.e. $\;$ $\left(y - 4\right) \left(y + 3\right) = 0$
i.e. $\;$ $y = 4$ $\;$ or $\;$ $y = -3$
When $\;$ $y = 4$, $\;$ we have from equation $(3)$
$x = 3^{4 - 1} = 3^3 = 27$
When $\;$ $y = -3$, $\;$ we have from equation $(3)$
$x = 3^{-3 - 1} = 3^{-4} = \dfrac{1}{81}$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(27, \; 4\right), \; \left(\dfrac{1}{81}, \; -3\right) \right\}$