Arithmetic Progression with Logarithms

By change of base formula, $\frac{\log_3(2^x - 9)}{\log_3 4} = \log_4(2^x - 9)$ and $\frac{\log_5\left(2^x + \frac{17}{2}\right)}{\log_5 4} = \log_4\left(2^x + \frac{17}{2}\right)$. Also, $\frac{1}{2} = \log_4 2$.

Since the three terms are in Arithmetic Progression: $2 \log_4 (2^x - 9) = \log_4 2 + \log_4 \left(2^x + \frac{17}{2}\right)$ $\log_4 (2^x - 9)^2 = \log_4 \left(2 \cdot 2^x + 17\right)$ $(2^x - 9)^2 = 2 \cdot 2^x + 17$

Let $y = 2^x$ (where $y > 9$ for the logarithm to be defined): $(y - 9)^2 = 2y + 17$ $y^2 - 18y + 81 = 2y + 17$ $y^2 - 20y + 64 = 0$ $(y - 16)(y - 4) = 0$

Since $y > 9$, we have $y = 16$, which gives $2^x = 16$.

The second term is $\log_4 (16 - 9) = \log_4 7$. The first term is $\log_4 2$.

Common difference $d = \log_4 7 - \log_4 2 = \log_4 \left(\frac{7}{2}\right)$.