Solve the equation: $\;$ $2^{2x-3} = 4^{x^2 - 3x -1}$
Given equation: $\;\;$ $2^{2x-3} = 4^{x^2 - 3x -1}$
i.e. $\;$ $2^{2x-3} = \left(2^2\right)^{x^2 - 3x -1}$
i.e. $\;$ $2^{2x - 3} = 2^{2x^2 - 6x - 2}$
$\implies$ $2x - 3 = 2x^2 - 6x - 2$
i.e. $\;$ $2x^2 - 8x + 1 = 0$
i.e. $\;$ $x = \dfrac{8 \pm \sqrt{64 - 8}}{4} = \dfrac{8 \pm \sqrt{56}}{4} = \dfrac{8 \pm 2 \sqrt{14}}{4}$
i.e. $\;$ $x = 2 \pm \dfrac{\sqrt{14}}{2} = 2 \pm \sqrt{\dfrac{7}{2}}$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = 2 \pm \sqrt{\dfrac{7}{2}}$