For a real number $x$, if $\frac{1}{2}, \frac{\log _{3}\left(2^{x}-9\right)}{\log _{3} 4}$, and $\frac{\log _{5}\left(2^{x}+\frac{17}{2}\right)}{\log _{5} 4}$ are in an arithmetic progression, then the common difference is

We have three terms in arithmetic progression (AP): $\frac{1}{2}$, $\frac{\log_3(2^x-9)}{\log_3 4}$, and $\frac{\log_5(2^x+\frac{17}{2})}{\log_5 4}$

If three terms a, b, c are in AP, then $2b = a + c$ (the middle term is the average of the first and third terms)

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The change of base formula states: $\frac{\log_a b}{\log_a c} = \log_c b$

Applying this:

$\frac{\log_3(2^x-9)}{\log_3 4} = \log_4(2^x-9)$

$\frac{\log_5(2^x+\frac{17}{2})}{\log_5 4} = \log_4(2^x+\frac{17}{2})$

This converts both logarithms to base 4, making our calculations much easier.

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We need to express $\frac{1}{2}$ as a logarithm with base 4.

Since $4^{1/2} = 2$, we have: $\frac{1}{2} = \log_4 2$

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Our three terms are now: $\log_4 2$, $\log_4(2^x-9)$, $\log_4(2^x+\frac{17}{2})$

For AP: $2 \times \log_4(2^x-9) = \log_4 2 + \log_4(2^x+\frac{17}{2})$

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Using $2\log_a b = \log_a b^2$ and $\log_a b + \log_a c = \log_a(bc)$:

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

Since the bases are equal, we can equate the arguments:

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

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Let $y = 2^x$ (this substitution makes the algebra much cleaner):

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

$y^2 - 18y + 81 = 2y + 17$

$y^2 - 20y + 64 = 0$

$(y-4)(y-16) = 0$

So $y = 4$ or $y = 16$

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Since $y = 2^x$:

If $y = 4$, then $2^x = 4 = 2^2$, so $x = 2$

If $y = 16$, then $2^x = 16 = 2^4$, so $x = 4$

For $x = 2$: $2^x - 9 = 4 - 9 = -5 < 0$

Since we can't take the logarithm of a negative number, $x = 2$ is invalid.

Therefore, $x = 4$ is our solution.

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With $x = 4$:

First term: $\log_4 2$

Second term: $\log_4(2^4-9) = \log_4(16-9) = \log_4 7$

Third term: $\log_4(2^4+\frac{17}{2}) = \log_4(16+8.5) = \log_4(\frac{49}{2})$

Common difference = Second term - First term = $\log_4 7 - \log_4 2 = \log_4(\frac{7}{2})$

The answer is $\log_4(\frac{7}{2})$