For a real number $x$, if $\frac{1}{2}$, $\frac{{\log }_3(2^x-9)}{{\log }_{3}4}$, and $\frac{{\log }_5(2^x+\frac{17}{2})}{{\log }_{5}4}$ are in an arithmetic progression, then the common difference is

The correct answer is ${\log }_4\left(\frac{7}{2}\right)$.

Let the three terms of the arithmetic progression be:
$a_1=\frac{1}{2}$
$a_2=\frac{{\log }_3(2^x-9)}{{\log }_{3}4}$
$a_3=\frac{{\log }_5(2^x+\frac{17}{2})}{{\log }_{5}4}$

First, we can simplify $a_2$ and $a_3$ using the change of base formula, $\frac{{\log }_{b}A}{{\log }_{b}B}={\log }_{B}A$:
$a_2={\log }_4(2^x-9)$
$a_3={\log }_4(2^x+\frac{17}{2})$

We also express $a_1=\frac{1}{2}$ with base 4:
$a_1={\log }_4\left(4^{1/2}\right)={\log }_{4}2$

Since $a_1$, $a_2$, and $a_3$ are in arithmetic progression, the middle term is the average of the first and third terms:
$2a_2=a_1+a_3$

Substituting the logarithmic expressions:
$2{\log }_4(2^x-9)={\log }_{4}2+{\log }_4(2^x+\frac{17}{2})$

Using properties of logarithms, $n\log A=\log \left(A^n\right)$ and $\log A+\log B=\log \left(AB\right)$:
${\log }_4{(2^x-9)}^2={\log }_4\left[2\right(2^x+\frac{17}{2}\left)\right]$

Equating the arguments:
${(2^x-9)}^2=2\left(2^x\right)+17$

Let $y=2^x$. Note that since the argument of ${\log }_3(2^x-9)$ must be positive, we require $2^x>9$, so $y>9$.
The equation becomes:
${(y-9)}^2=2y+17$
$y^2-18y+81=2y+17$
$y^2-20y+64=0$

Factoring the quadratic equation:
$(y-16)(y-4)=0$
This gives $y=16$ or $y=4$.

Since we must satisfy $y>9$, we reject $y=4$. Thus:
$y=2^x=16$

Now we calculate the common difference $d$ of the arithmetic progression:
$d=a_2-a_1$
$d={\log }_4(2^x-9)-\frac{1}{2}$
$d={\log }_4(16-9)-{\log }_{4}2$
$d={\log }_{4}7-{\log }_{4}2$
$d={\log }_4\left(\frac{7}{2}\right)$