The value of x in the given equation $4^x-3^{x-\frac{1}{2}}=3^{x+\frac{1}{2}}-2^{2x-1}$ is

Equation, $4^x-3^{x-\frac{1}{2}}=3^{x+\frac{1}{2}}-2^{2x-1}$

$\Rightarrow 2^{2x}+2^{2x-1}=3^{x+\frac{1}{2}}+3^{x-\frac{1}{2}}$

$\Rightarrow 2^{2x}\left(1+\frac{{\displaystyle 1}}{{\displaystyle 2}}\right)=3^{x-\frac{1}{2}}(1+3)$

$\Rightarrow 2^{2x}.\frac{3}{2}=3^{x-\frac{1}{2}}.4 \Rightarrow 2^{2x-3}=3^{x-\frac{3}{2}}$

Taking log both sides

$\Rightarrow (2x-3)\log 2=(x-3/2)\log 3$

$\Rightarrow 2x\log 2-3\log 2=x\log 3-\frac{3}{2}\log 3$

$\Rightarrow x\log 4-x\log 3=3\log 2-\frac{3}{2}\log 3$

$\Rightarrow x\log \left(\frac{{\displaystyle 4}}{{\displaystyle 3}}\right)=\log 8-\log 3\sqrt{3}$

$\Rightarrow {\left(\frac{{\displaystyle 4}}{{\displaystyle 3}}\right)}^x=\frac{8}{3\sqrt{3}} \Rightarrow {\left(\frac{{\displaystyle 4}}{{\displaystyle 3}}\right)}^x={\left(\frac{{\displaystyle 4}}{{\displaystyle 3}}\right)}^{3/2}$

$\therefore x=\frac{3}{2}$

Trick: Cheak the equation with options then only option (2) satisfies the equation.