Question

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

• A. \(\sqrt{a}\)
• B. 2a
• C. \(\frac{a}{2}\)
• D. a

Answer 1

The Correct Option is D

Given \(log_3 \ x = log_{12} \ y = a\), let us express \(x\) and \(y\) in terms of \(a\).

1. From \(log_3 \ x = a\), we have \(x = 3^a\).

2. From \(log_{12} \ y = a\), we have \(y = 12^a\).

The geometric mean \(G\) of \(x\) and \(y\) is given by:

\[G = \sqrt{xy} = \sqrt{3^a \cdot 12^a} = \sqrt{(3 \cdot 12)^a} = \sqrt{36^a} = (36^a)^{1/2} = 36^{a/2}\]

We are asked to find \(log_6 \ G\).

Substitute \(G = 36^{a/2}\) into the logarithmic expression:

\[log_6 \ G = log_6 (36^{a/2}) = \frac{a}{2} \cdot log_6 \ 36\]

Calculate \(log_6 \ 36\):

Since \(36 = 6^2\), we have:

\[log_6 \ 36 = log_6 (6^2) = 2 \cdot log_6 \ 6 = 2 \cdot 1 = 2\]

Substitute back to find \(log_6 \ G\):

\[log_6 \ G = \frac{a}{2} \cdot 2 = a\]

Therefore, \(log_6 \ G = a\).