3 3 votes 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 $\sqrt{a}$ $2a$ $a/2$ $a$ Quantitative Aptitude cat2017-1 quantitative-aptitude geometric-progression geometric-mean + – go_editor 14.2k points 2.5k views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
10 10 votes 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}$ $\textbf{PS:}$ $\log_{b}{a^{x}} = x\log_{b}{a}$ $\log_{a}{a} = 1$ Anjana5051 answered Dec 27, 2021 • edited Dec 27, 2021 by Lakshman Bhaiya Anjana5051 12.1k points comment Share Follow 0 reply Please log in or register to add a comment.