Step 1: Write the equation clearly.
We need to solve $4^x - 3^{x-\frac{1}{2}} = 3^{x+\frac{1}{2}} - 2^{2x-1}$. Notice that $4^x = 2^{2x}$ and $2^{2x-1} = \frac{2^{2x}}{2}$.
Step 2: Rewrite all terms with consistent bases.
$3^{x-\frac{1}{2}} = \dfrac{3^x}{\sqrt{3}}$ and $3^{x+\frac{1}{2}} = 3^x\sqrt{3}$. The equation becomes: \[2^{2x} - \frac{3^x}{\sqrt{3}} = 3^x\sqrt{3} - \frac{2^{2x}}{2}.\]
Step 3: Group like terms.
Move all $2^{2x}$ terms to the left and all $3^x$ terms to the right: \[2^{2x} + \frac{2^{2x}}{2} = 3^x\sqrt{3} + \frac{3^x}{\sqrt{3}}.\] \[2^{2x}\left(1 + \frac{1}{2}\right) = 3^x\left(\sqrt{3} + \frac{1}{\sqrt{3}}\right).\] \[\frac{3}{2} \cdot 2^{2x} = 3^x \cdot \frac{3+1}{\sqrt{3}} = \frac{4 \cdot 3^x}{\sqrt{3}}.\]
Step 4: Simplify and take the ratio.
\[\frac{3}{2} \cdot 4^x = \frac{4 \cdot 3^x}{\sqrt{3}} \implies \frac{4^x}{3^x} = \frac{4}{\sqrt{3}} \cdot \frac{2}{3} = \frac{8}{3\sqrt{3}} = \frac{8}{3^{3/2}}.\] So $\left(\dfrac{4}{3}\right)^x = \dfrac{8}{3\sqrt{3}} = \dfrac{2^3}{3^{3/2}} = \left(\dfrac{4}{3}\right)^{3/2}$.
Step 5: Equate exponents.
$x = \dfrac{3}{2}$.
Step 6: Verify the solution.
At $x=\frac{3}{2}$: $4^{3/2}=8$, $3^1 = 3$ (for $3^{x-1/2}$), $3^2=9$ (for $3^{x+1/2}$), $2^2=4$ (for $2^{2x-1}$). Check: $8-3=5$ and $9-4=5$. ✓ \[ \boxed{x = \dfrac{3}{2}} \]