Solve the equation: $\;$ $\log_3 \left(3^x - 6\right) = x - 1$
Given equation: $\;\;$ $\log_3 \left(3^x - 6\right) = x - 1$
i.e. $\;$ $3^x - 6 = 3^{x-1}$
i.e. $\;$ $3^x - 6 = 3^x \times 3^{-1}$
i.e. $\;$ $3^x - 6 = \dfrac{3^x}{3}$
i.e. $\;$ $3^x \left(1 - \dfrac{1}{3}\right) = 6$
i.e. $\;$ $3^x \times \dfrac{2}{3} = 6$
i.e. $\;$ $3^{x-1} = 3 = 3^1$
$\implies$ $x - 1 = 1$ $\implies$ $x = 2$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{2 \right\}$