If $a^{x} = b c, b^{y} = c a$ and $c^{z} = a b$ , show that $x y z = x + y + z + 2$

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Solution

$a^{x} = b c$
$\Rightarrow {(a^{x})}^{y} = {(b c)}^{y}$
$\Rightarrow a^{x y} = b^{y} c^{y}$
$\Rightarrow a^{x y} = (c a) (c^{y})$
$\Rightarrow {(a^{x y})}^{z} = {(c a)}^{z} {(c^{y})}^{z}$
$\Rightarrow (a^{x y z}) = c^{z} a^{z} {(c^{z})}^{y}$
$\Rightarrow (a^{x y z}) = (a b) a^{z} {(a b)}^{y}$
$\Rightarrow (a^{x y z}) = (a b) a^{z} a^{y} b^{y}$
$\Rightarrow (a^{x y z}) = (a b) a^{z} a^{y} (c a)$
$\Rightarrow (a^{x y z}) = a^{2 + y + z} (b c)$
$\Rightarrow (a^{x y z}) = a^{2 + y + z} a^{x}$
$\Rightarrow (a^{x y z}) = a^{2 + x + y + z}$
$\Rightarrow (a^{x y z}) = a^{2 + x + y + z}$
Since bases are same, we can equate the powers,
$∴ x y z = x + y + z + 2$
Hence proved.

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