If log_x y, log_z x, log_y z are in G.P.,xyz | vedclass

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(D) Given $\log_x y, \log_z x, \log_y z$ are in $G.P.$Therefore,$(\log_z x)^2 = (\log_x y)(\log_y z) = \log_x z = \frac{1}{\log_z x}$.This implies $(\

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(D) Given $\log_x y, \log_z x, \log_y z$ are in $G.P.$Therefore,$(\log_z x)^2 = (\log_x y)(\log_y z) = \log_x z = \frac{1}{\log_z x}$.This implies $(\log_z x)^3 = 1$,so $\log_z x = 1$,which means $x = z$.Given $xyz = 64$ and $x = z$,we have $x^2 y = 64$,so $y = \frac{64}{x^2}$.Given $x^3, y^3, z^3$ are in $A.P.$,we have $2y^3 = x^3 + z^3$.Since $x = z$,$2y^3 = 2x^3$,which implies $y^3 = x^3$,so $y = x$.Substituting $y = x$ into $x^2 y = 64$,we get $x^3 = 64$,so $x = 4$.Thus,$x = y = z = 4$.Since $x = y = z = 4$,all conditions are satisfied: $x=y=z$ is true,$x=4$ is true,and $x, y, z$ are in $G.P.$ (with common ratio $1$) is true.Therefore,the correct option is $(d)$.

If $\log_x y, \log_z x, \log_y z$ are in $G.P.$,$xyz = 64$,and $x^3, y^3, z^3$ are in $A.P.$,then

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