Solve the equation: $\;$ $3^{12x-1} - 9^{6x-1} - 27^{4x-1} + 81^{3x+1} = 2192$
Given equation: $\;\;$ $3^{12x-1} - 9^{6x-1} - 27^{4x-1} + 81^{3x+1} = 2192$
i.e. $\;$ $3^{12x-1} - \left(3^2\right)^{6x-1} - \left(3^3\right)^{4x-1} + \left(3^4\right)^{3x + 1} = 2192$
i.e. $\;$ $3^{12x-1} - 3^{12x-2} - 3^{12x-3} + 3^{12x+4} = 2192$
i.e. $\;$ $3^{12x} \left(3^{-1} - 3^{-2} - 3^{-3} + 3^4\right) = 2192$
i.e. $\;$ $3^{12x} \left(\dfrac{1}{3} - \dfrac{1}{3^2} - \dfrac{1}{3^3} + 3^4\right) = 2192$
i.e. $\;$ $3^{12x} \left(\dfrac{3^2 - 3 - 1}{3^3} + 3^4\right) = 2192$
i.e. $\;$ $3^{12x} \times \dfrac{2192}{27} = 2192$
i.e. $\;$ $3^{12x} = 27 = 3^3$
$\implies$ $12x = 3$
i.e. $\;$ $x = \dfrac{3}{12} = \dfrac{1}{4}$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{1}{4} \right\}$