Solve the equation: $\;$ $3^{2x - 3} - 9^{x - 1} + 27^{\frac{2x}{3}} = 675$
Given equation: $\;\;$ $3^{2x - 3} - 9^{x - 1} + 27^{\frac{2x}{3}} = 675$
i.e. $\;$ $3^{2x} \times 3^{-3} - \left(3^2\right)^{x - 1} + \left(3^3\right)^{\frac{2x}{3}} = 25 \times 27$
i.e. $\;$ $3^{2x} \times 3^{-3} - 3^{2x} \times 3^{-2} + 3^{2x} = 25 \times 3^3$
i.e. $\;$ $3^{2x} \left(\dfrac{1}{3^3} - \dfrac{1}{3^2} + 1\right) = 25 \times 3^3$
i.e. $\;$ $3^{2x - 3} \left(\dfrac{1 - 3 + 27}{27}\right) = 25$
i.e. $\;$ $3^{2x - 3} \times \dfrac{25}{27} = 25$
i.e. $\;$ $3^{2x - 3} = 27 = 3^3$
$\implies$ $2x - 3 = 3$
i.e. $\;$ $2x = 6$ $\implies$ $x = 3$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = 3$