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(C) Given that $\log _{a} x, \log _{b} x, \log _{c} x$ are in $AP$.Therefore, $2 \log _{b} x = \log _{a} x + \log _{c} x$.Using the change of base for

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$c^{2}=(a c)^{\log_{a} b}$

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(C) Given that $\log _{a} x, \log _{b} x, \log _{c} x$ are in $AP$.Therefore, $2 \log _{b} x = \log _{a} x + \log _{c} x$.Using the change of base formula $\log _{m} n = \frac{1}{\log _{n} m}$, we get:$\frac{2}{\log _{x} b} = \frac{1}{\log _{x} a} + \frac{1}{\log _{x} c}$.$\frac{2}{\log _{x} b} = \frac{\log _{x} c + \log _{x} a}{\log _{x} a \cdot \log _{x} c} = \frac{\log _{x} (ac)}{\log _{x} a \cdot \log _{x} c}$.Rearranging, we have $\frac{2}{\log _{x} b} = \frac{\log _{x} (ac)}{\log _{x} a \cdot \log _{x} c}$.Since $\frac{1}{\log _{x} a} = \log _{a} x$, we can write $\frac{\log _{x} a}{\log _{x} b} = \log _{b} a$, so $\frac{1}{\log _{x} b} = \log _{b} x$.Actually, $\frac{2}{\log _{x} b} = 2 \log _{b} x$. Thus, $2 \log _{b} x = \frac{\log _{x} (ac)}{\log _{x} a \cdot \log _{x} c}$.Using $\log _{x} a \cdot \log _{a} b = \log _{x} b$, we get $\log _{x} b = \log _{x} a \cdot \log _{a} b$.Substituting this, $2 \log _{b} x = \frac{\log _{x} (ac)}{\log _{x} a \cdot \log _{x} c}$.This simplifies to $\log _{x} c^2 = \log _{x} (ac)^{\log _{a} b}$.Thus, $c^2 = (ac)^{\log _{a} b}$.

If $x$ is a positive real number different from $1$ such that $\log _{a} x, \log _{b} x, \log _{c} x$ are in $AP$, then

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