Given $x^{3}, y^{3}, z^{3}$ are in $A . P .$ and ${l o g}_{x} y, {l o g}_{z} x, {l o g}_{y} z$ are in $G . P .$
If $x y z = 64$ , then prove that $x = y = z = 4$ .

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Solution

${2 y}^{3} = x^{3} + z^{3}$ or ${(\frac{x}{y})}^{3} + {(\frac{z}{y})}^{3} = 2$ ...... $(1)$
${({l o g}_{z} x)}^{2} = {l o g}_{x} y . {l o g}_{y} z$
or ${(\frac{l o g x}{l o g z})}^{2} = \frac{l o g y}{l o g x}, \frac{l o g z}{l o g y} = \frac{l o g z}{l o g x}$
$∴ {(l o g x)}^{3} = {(l o g z)}^{3}$ or $l o g x = l o g z$
$∴ x = z$
Hence from $(1), {2 y}^{3} = {2 z}^{3} ∴ y = z = x$
But $x y z = 64 ∴ x^{3} = 64$
$∴ x = y = z = 4$

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