Solve the equation: $\;$ $2 \times 3^{x + 1} - 6 \times 3^{x-1} - 3^x = 9$
Given equation: $\;\;$ $2 \times 3^{x + 1} - 6 \times 3^{x-1} - 3^x = 9$
i.e. $\;$ $2 \times 3^x \times 3^1 - 6 \times 3^x \times 3^{-1} - 3^x = 3^2$
i.e. $\;$ $6 \times 3^x - 3 \times 3^x - 3^x = 3^2$
i.e. $\;$ $3 \times 3^x = 3^2$
i.e. $\;$ $3^{x+1} = 3^2$
$\implies$ $x + 1 = 2$ $\implies$ $x = 1$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{1 \right\}$