Question 87

Suppose, $$\log_3 x = \log_{12} y = a$$, where $$x, y$$ are positive numbers. If $$G$$ is the geometric mean of x and y, and $$\log_6 G$$ is equal to


We know that $$\log_3 x = a$$ and $$\log_{12} y=a$$
Hence, $$x = 3^a$$ and $$y=12^a$$
Therefore, the geometric mean of $$x$$ and $$y$$ equals $$\sqrt{x \times y}$$
This equals $$\sqrt{3^a \times 12^a} = 6^a$$

Hence, $$G=6^a$$ Or, $$\log_6 G = a$$

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