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Chain Rule of Differentiation

If x = log1...

Question

If $x = l o g_{12} 6, y = l o g_{18} 12$ and $z = l o g_{24} 18$ prove that $1 + x y z = 2 y z$ .

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Solution

Given: $x = l o g_{12} 6, y = l o g_{18} 12, l o g_{24} 18$
To prove: $1 + x y z - 2 y z$
$= 1 + l o g_{12} 6 l o g_{18} 12 l o g_{24} 18$
$= 1 + \frac{l o g e_{6}}{l o g_{e} 12} \frac{l o g e_{12}}{l o g e_{18}} . \frac{l o g e_{18}}{l o g e_{24}}$ [using $l o g a b = \frac{l o g_{e} b}{l o g_{e} b}]$
$= 1 + \frac{l o g_{e} 6}{l o g_{e} 24}$
$= 1 + l o g_{24} 6$
$= l o g_{24} 24 + l o g_{24} 6 [∵ l o g_{e} a = 1]$
$= l o g_{24} 144 [∴ l o g_{a} b + l o g_{a} C = l o g_{a} b c]$
$= 2 l o g_{24} 12$
$= \frac{2 l o g_{e}}{l o g_{e}} \frac{12}{18} . \frac{l o g_{e} 18}{l o g_{e} 24} = 2 y z = R H S$

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