CAT 2023

Slot 3

QA

Question & Solution

It is given that $\frac{1}{2}, \frac{\log_3(2^x - 9)}{\log_3 4}$, and $\frac{\log_5\left(2^x + \frac{17}{2}\right)}{\log_5 4}$ are in an arithmetic progression.

$\frac{\log_3(2^x-9)}{\log_34}$ can be written as $\log_4\left(2^x-9\right)$, and $\frac{\log_5\left(2^x + \frac{17}{2}\right)}{\log_5 4}$ can be written as $\log_4\left(2^x+\frac{17}{2}\right)$ 

Hence, $2\log_4\left(2^x-9\right)=\frac{1}{2}+\log_4\left(2^x+\frac{17}{2}\right)$

$\frac{1}{2}$ can be written as $\log_42$.

Therefore, 

=> $2\log_4\left(2^x-9\right)=\log_42+\log_4\left(2^x+\frac{17}{2}\right)$

=> $\log_4\left(2^x-9\right)^{^2}=\log_42\left(2^x+\frac{17}{2}\right)$

=> $\left(2^x-9\right)^{^2}=2\left(2^x+\frac{17}{2}\right)$

=> $2^{2x}-18\cdot2^x+81=2\cdot2^x+17$

=> $2^{2x}-20\cdot2^x+64=0$

=> $2^{2x}-16\cdot2^x-4\cdot2^x+64=0$

=> $2^x\left(2^x-16\right)-4\left(2^x-16\right)=0$

=> $\left(2^x-4\right)\left(2^x-16\right)=0$

The values of $2^x$ can't be 4 (log will be undefined), which implies The value of $2^x$ is 16.

Therefore, the common difference is $\log_4\left(2^x-9\right)-\log_42$

=> $\log_47-\log_42\ =\ \log_4\left(\frac{7}{2}\right)$

The correct option is D