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