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Question 21 : 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
- 2a
- \\frac{a}{2})
- a

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log_{3} x = log_{12} y = a, where x, y are positive numbers.

log_{3} x = a; x = 3^{a}

log_{12} y = a; y = 12^{a}

If G is the geometric mean of x and y, log6G is equal to has to be found

We know that geometric mean is √xy

√xy = √(36^{a})

= √(6^{2a}) = (6^{2a})^{\\frac{1}{2})}

= 6^{a}

log_{6} G = a

The question is **"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 :" **

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