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 have: $\log_3 x = \log_{12} y = a$

This means both logarithmic expressions equal the same value $a$.

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If $\log_b m = n$, then $b^n = m$

From $\log_3 x = a$:

$x = 3^a$

From $\log_{12} y = a$:

$y = 12^a$

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Now we can find $xy$:

$xy = 3^a \times 12^a$

Using the exponent rule $a^m \times b^m = (ab)^m$:

$xy = 3^a \times 12^a = (3 \times 12)^a = 36^a$

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The geometric mean of two positive numbers is the square root of their product.

For numbers $x$ and $y$: $G = \sqrt{xy}$

$G = \sqrt{xy} = \sqrt{36^a}$

Since $\sqrt{n} = n^{1/2}$:

$G = \sqrt{36^a} = (36^a)^{1/2}$

Using the exponent rule $(a^m)^n = a^{mn}$:

$G = (36^a)^{1/2} = 36^{a/2}$

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Notice that $36 = 6^2$, so:

$G = 36^{a/2} = (6^2)^{a/2}$

$G = (6^2)^{a/2} = 6^{2 \times a/2} = 6^a$

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We need to find $\log_6 G$ where $G = 6^a$.

Using the property $\log_b(b^x) = x$:

$\log_6 G = \log_6(6^a) = a$

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$\log_6 G = a$

The beauty of this problem is that despite the different bases (3, 12, and 6), the common value $a$ ties everything together through the properties of logarithms and exponents.