Question 1.

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

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

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

$\frac{1}{2}$ can be written as $\log_4 2$.

Therefore,

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

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

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

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

=> $2^{2x} - 20 \cdot 2^x + 64 = 0$

=> $2^{2x} - 16 \cdot 2^x - 4 \cdot 2^x + 64 = 0$

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

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

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 $\log_4(2^x - 9) - \log_4 2$

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