Given,
$4^x-3^{x-\frac{1}{2}}=3^{x+\frac{1}{2}}-2^{2 x-1}$
$\begin{aligned} & \Rightarrow \quad(2)^{2 x}+2^{2 x-1}=3^{x+\frac{1}{2}}+3^{x-\frac{1}{2}} \\ & \Rightarrow \quad 2^{2 x}+\frac{2^{2 x}}{2}=3^x \cdot \sqrt{3}+\frac{3^x}{\sqrt{3}} \\ & \Rightarrow \quad 2^{2 x}\left[1+\frac{1}{2}\right]=3^x\left[\sqrt{3}+\frac{1}{\sqrt{3}}\right]\end{aligned}$
$\begin{aligned} & \Rightarrow \quad 2^{2 x} \times \frac{3}{2}=3^x \times \frac{4}{\sqrt{3}} \\ & \Rightarrow \quad 2^{2 x}(3 \sqrt{3})=3^x(8) \\ & \Rightarrow \quad 2^{2 x}(3)^{\frac{3}{2}}=3^x \times(2)^3\end{aligned}$
Comparing power of $2,2 x=3$
and comparing power of $3, x=\frac{3}{2}$
Both equations given us $x=\frac{3}{2}$.