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Suppose, $\log_{3}x= \log_{12}y= a$, where $x, y$ are positive numbers. If $\text{G}$ is the geometric mean of $x$ and $y$, and $\log_{6}\text{G}$ is equal to

  1. $\sqrt{a}$
  2. $2a$
  3. $a/2$
  4. $a$
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Given that , $\log_{3}{x} = \log_{12}{y} = a \quad \longrightarrow (1) $   

From equation $(1),\;\log_{3}{x} = a$

$\Rightarrow \boxed{x = 3^{a}} \quad [\because \log_{a}{x} = b \Rightarrow x = a^{b}]$

Again, from equation $(1),\;\log_{12}{y} = a$

$\Rightarrow \boxed{y = 12^{a}}$

If $G$ is the geometric mean of $x$ and $y$, then $G= \sqrt{xy}$

$\Rightarrow G = \sqrt{3^{a}\cdot 12^{a}}  $

 $\Rightarrow G = \sqrt{3^{a}\cdot (3 \cdot 4)^{a}}$

$\Rightarrow G= \sqrt{3^{a}\cdot 3^{a}\cdot4^{a}}$

$\Rightarrow G= \sqrt{3^{2a}\cdot 2^{2a}}$

$\Rightarrow G= 3^{a}\cdot 2^{a}$

$\Rightarrow \boxed{G= 6^{a}}$

$\therefore$  The value of  $\log_{6}{G} = \log_{6}{6^{a}}  = a \log_{6}{6} = a$

Correct Answer $ : \text{D}$

$\textbf{PS:}$

  • $\log_{b}{a^{x}} = x\log_{b}{a}$
  • $\log_{a}{a} = 1$
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