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If $x$ is a positive real number different from 1 such that $\log _{a} x, \log _{b} x, \log _{c} x$ are in $\mathrm{AP}$, then
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The correct answer is:
$c^{2}=(a c)^{log_{a} b}$
Given, $\log _{a} x, \log _{b} x$ and $log_{c}$ $x$ are in $A P$.
$\therefore \quad 2 \log _{b} x=\log _{a} x+\log _{c} x$
$\begin{aligned} 2 \log _{b} x &=\frac{1}{\log _{x} a}+\frac{1}{\log _{x} c} \\ \frac{2}{\log _{x} b} &=\frac{\log _{x} a c}{\log _{x} a \log _{x} c} \\ 2 \log _{x} c &=\frac{\log _{x} b}{\log _{x} a}\left(\log _{x} a c\right) \\ 2 \log _{x} c &=\log _{a} b \cdot \log _{x} a c \\ \log _{x} c^{2} &=\log _{x}(a c)^{\log _{a} b} \\ c^{2} &=(a c)^{\log _{a} b} \end{aligned}$
$\therefore \quad 2 \log _{b} x=\log _{a} x+\log _{c} x$
$\begin{aligned} 2 \log _{b} x &=\frac{1}{\log _{x} a}+\frac{1}{\log _{x} c} \\ \frac{2}{\log _{x} b} &=\frac{\log _{x} a c}{\log _{x} a \log _{x} c} \\ 2 \log _{x} c &=\frac{\log _{x} b}{\log _{x} a}\left(\log _{x} a c\right) \\ 2 \log _{x} c &=\log _{a} b \cdot \log _{x} a c \\ \log _{x} c^{2} &=\log _{x}(a c)^{\log _{a} b} \\ c^{2} &=(a c)^{\log _{a} b} \end{aligned}$
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