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Given that, $9^{\left(x−\frac{1}{2}\right)}−2^{(2x−2)}=4^{x}−3^{(2x−3)}$

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

$\Rightarrow 3^{(2x-1)}-2^{(2x−2)}=2^{2x}−3^{(2x−3)}$

$\Rightarrow (3^{2x-1})+3^{(2x−3)}=2^{2x}+2^{(2x−2)}$

$\Rightarrow 3^{2x}\cdot3^{−1}+3^{2x}\cdot3^{−3}=2^{2x}+2^{2x}\cdot2^{−2}$

$\Rightarrow \dfrac{3^{2x}}{3}+\dfrac{3^{2x}}{27}=2^{2x}+\dfrac{2^{2x}}{4}$

$\Rightarrow \dfrac{9 \cdot 3^{2x}+3^{2x}}{27}=\dfrac{4 \cdot2^{2x}+2^{2x}}{4}$

$\Rightarrow \dfrac{3^{2x}(9+1)}{27}=\dfrac{2^{2x}(4+1)}{4} $

$\Rightarrow \dfrac{10}{27}(3^{2x})=\dfrac{5}{4}(2^{2x})$

$\Rightarrow \dfrac{3^{2x}}{27}=\dfrac{2^{2x}}{8}$

$\Rightarrow 3^{2x} \cdot 3^{-3} = 2^{2x} \cdot 2^{-3}$

$\Rightarrow \boxed{ 3^{(2x-3)}=2^{(2x-3)}}$

This is only possible when, $2x-3=0$

$\Rightarrow \boxed{x=\frac{3}{2}}$

Correct Answer $ :\text{A}$

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

$\Rightarrow 3^{(2x-1)}-2^{(2x−2)}=2^{2x}−3^{(2x−3)}$

$\Rightarrow (3^{2x-1})+3^{(2x−3)}=2^{2x}+2^{(2x−2)}$

$\Rightarrow 3^{2x}\cdot3^{−1}+3^{2x}\cdot3^{−3}=2^{2x}+2^{2x}\cdot2^{−2}$

$\Rightarrow \dfrac{3^{2x}}{3}+\dfrac{3^{2x}}{27}=2^{2x}+\dfrac{2^{2x}}{4}$

$\Rightarrow \dfrac{9 \cdot 3^{2x}+3^{2x}}{27}=\dfrac{4 \cdot2^{2x}+2^{2x}}{4}$

$\Rightarrow \dfrac{3^{2x}(9+1)}{27}=\dfrac{2^{2x}(4+1)}{4} $

$\Rightarrow \dfrac{10}{27}(3^{2x})=\dfrac{5}{4}(2^{2x})$

$\Rightarrow \dfrac{3^{2x}}{27}=\dfrac{2^{2x}}{8}$

$\Rightarrow 3^{2x} \cdot 3^{-3} = 2^{2x} \cdot 2^{-3}$

$\Rightarrow \boxed{ 3^{(2x-3)}=2^{(2x-3)}}$

This is only possible when, $2x-3=0$

$\Rightarrow \boxed{x=\frac{3}{2}}$

Correct Answer $ :\text{A}$