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