If $\log _{4} 5=\left(\log _{4} y\right)\left(\log _{6} \sqrt{ } 5\right)$, then $y$ equals

Given: $\log_4 5 = (\log_4 y)(\log_6 \sqrt{5})$

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$\log_6 \sqrt{5} = \log_6 5^{1/2} = \dfrac{1}{2}\log_6 5$

The logarithm of a square root equals half the logarithm of the number inside, using the power rule: $\log_a b^n = n \log_a b$.

Our equation becomes:

$\log_4 5 = (\log_4 y) \cdot \dfrac{1}{2}\log_6 5$

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$\dfrac{\log_4 5}{\dfrac{1}{2}\log_6 5} = \log_4 y$

This simplifies to:

$\dfrac{2\log_4 5}{\log_6 5} = \log_4 y$

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Using the change of base formula: $\log_a b = \dfrac{\log b}{\log a}$

$\log_4 5 = \dfrac{\log 5}{\log 4}$

$\log_6 5 = \dfrac{\log 5}{\log 6}$

Substituting:

$\dfrac{2 \cdot \dfrac{\log 5}{\log 4}}{\dfrac{\log 5}{\log 6}} = \log_4 y$

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$\dfrac{2\log 5}{\log 4} \cdot \dfrac{\log 6}{\log 5} = \log_4 y$

The $\log 5$ terms cancel:

$\dfrac{2\log 6}{\log 4} = \log_4 y$

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Using the change of base formula in reverse:

$\dfrac{2\log 6}{\log 4} = 2 \cdot \dfrac{\log 6}{\log 4} = 2\log_4 6$

Therefore: $2\log_4 6 = \log_4 y$

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Using the power rule ($n\log_a b = \log_a b^n$):

$\log_4 6^2 = \log_4 y$

Therefore: $y = 36$