CAT 2023 Slot 3 Solved Paper

It is given that $\frac{1}{2},\frac{{\log }_3\left(2^{\mathit{x}}-9\right)}{{\log }_{3}4}$, and $\frac{{\log }_5\left(2^{\mathit{x}}+\frac{17}{2}\right)}{{\log }_{5}4}$ are in an arithmetic progression.
$\frac{{\log }_3\left(2^{\mathit{x}}-9\right)}{{\log }_{3}4}$ can be written as ${\log }_4\left(2^{\mathit{x}}-9\right)$, and $\frac{{\log }_5\left(2^{\mathit{x}}+\frac{17}{2}\right)}{{\log }_{5}4}$ can be written as ${\log }_4\left(2^{\mathit{x}}+\frac{17}{2}\right)$
Hence, $2{\log }_4\left(2^{\mathit{x}}-9\right)=\frac{1}{2}+{\log }_4\left(2^{\mathit{x}}+\frac{17}{2}\right)$
$\frac{1}{2}$ can be written as ${\log }_{4}2$.
Therefore,
$\Rightarrow 2{\log }_4\left(2^{\mathit{x}}-9\right)={\log }_{4}2+{\log }_4\left(2^{\mathit{x}}+\frac{17}{2}\right)$
$\Rightarrow {\log }_4{\left(2^{\mathit{x}}-9\right)}^2={\log }_{4}2\left(2^{\mathit{x}}+\frac{17}{2}\right)$
$\Rightarrow {\left(2^{\mathit{x}}-9\right)}^2=2\left(2^{\mathit{x}}+\frac{17}{2}\right)$
$\Rightarrow 2^{2\mathit{x}}-18\cdot 2^{\mathit{x}}+81=2\cdot 2^{\mathit{x}}+17$
$\Rightarrow 2^{2\mathit{x}}-20\cdot 2^{\mathit{x}}+64=0$
$\Rightarrow 2^{2\mathit{x}}-16\cdot 2^{\mathit{x}}-4\cdot 2^{\mathit{x}}+64=0$
$\Rightarrow 2^{\mathit{x}}\left(2^{\mathit{x}}-16\right)-4\left(2^{\mathit{x}}-16\right)=0$
$\Rightarrow \left(2^{\mathit{x}}-4\right)\left(2^{\mathit{x}}-16\right)=0$
The values of $2^{\wedge }\mathit{x}\mathit{x}$ can't be 4 (log will be undefined), which implies The value of $2^{\mathit{x}}$ is 16.
Therefore, the common difference is ${\log }_4\left(2^{\mathit{x}}-9\mathit{t}\right)-{\log }_{4}2$
$\Rightarrow {\log }_{4}7-{\log }_{4}2={\log }_4\left(\frac{7}{2}\right)$
The correct option is $\mathit{D}$