If log_ax,log_bx log_cx are in A.P. where  x not equal to 1, show that c²= (ac)^log _a ^b 

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ {\log }_{\mathrm{a}}\mathrm{x},{\log }_{\mathrm{b}}\mathrm{x}, {\log }_{\mathrm{c}}\mathrm{x} \mathrm{in} \mathrm{A}.\mathrm{P}. \\ \mathrm{Hence} \\ 2{\log }_{\mathrm{b}}\mathrm{x}={\log }_{\mathrm{a}}\mathrm{x}+ {\log }_{\mathrm{c}}\mathrm{x} \\ \Rightarrow \frac{2}{{\log }_{\mathrm{x}}\mathrm{b}}=\frac{1}{{\log }_{\mathrm{x}}\mathrm{a}}+\frac{1}{{\log }_{\mathrm{x}}\mathrm{c}} \\ \Rightarrow \frac{2}{{\log }_{\mathrm{x}}\mathrm{b}}=\frac{{\log }_{\mathrm{x}}\mathrm{a}+{\log }_{\mathrm{x}}\mathrm{c}}{{\log }_{\mathrm{x}}\mathrm{a}.{\log }_{\mathrm{x}}\mathrm{c}} \\ \Rightarrow 2{\log }_{\mathrm{x}}\mathrm{c}=\frac{{\log }_{\mathrm{x}}\mathrm{b}}{{\log }_{\mathrm{x}}\mathrm{a}}\left({\log }_{\mathrm{x}}\left(\mathrm{ac}\right)\right) \\ \Rightarrow {\log }_{\mathrm{x}}{\mathrm{c}}^2={\log }_{\mathrm{a}}\mathrm{b}\left({\log }_{\mathrm{x}}\left(\mathrm{ac}\right)\right) \\ \Rightarrow {\log }_{\mathrm{x}}{\mathrm{c}}^2=\left({\log }_{\mathrm{x}}{\left(\mathrm{ac}\right)}^{{\log }_{\mathrm{a}}\mathrm{b}}\right) \\ \Rightarrow \boxed{{\mathbf{c}}^{\mathbf{2}}\mathbf{=}{\left(\mathbf{ac}\right)}^{{\mathbf{log}}_{\mathbf{a}}\mathbf{b}}} \end{gathered}$$

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