Given that, $ \log_{4}5 = (\log_{4}y) (\log_{6}\sqrt{5}) $
$ \Rightarrow \dfrac{ \log_{4}5}{ \log_{4}y} = \log_{6} \sqrt{5} $
$ \Rightarrow \log_{y}5 = \log_{6} \sqrt{5} \quad \left[ \because \log_{a}b = \frac{\log_{x}b}{\log_{x}a}\right] $
Let $ \log_{6} \sqrt{5} = k $
$ \Rightarrow \sqrt{5} = 6^{k} \quad [ \because \log_{a}b = x \Rightarrow b = a^{x}] $
On squaring both sides.
$ \Rightarrow ( \sqrt{5})^{2} = (6^{k})^{2} $
$ \Rightarrow 5 = 6^{2k} $
$ \Rightarrow \boxed{36^{k} = 5} $
Now, $ \log_{y}5 = k $
$ \Rightarrow 5 = y^{k} $
$ \Rightarrow y^{k} = 36^{k} \quad [ \because 5 = 36^{k}] $
$ \Rightarrow \boxed {y = 36} $
$\textbf{Second Method:}$
Given that, $\log_{4}5 = (\log_{4}y)(\log_{6} \sqrt{5}) $
$ \Rightarrow \log_{4}y = \dfrac{ \log_{4}5}{\log_{6}5^{\frac{1}{2}}} $
$ \Rightarrow \log_{4}y = \dfrac{\log_{4}5}{\frac{1}{2} \log_{6}5} \quad [ \because \log_{b}a^{x} = x \log_{b}a] $
$ \Rightarrow \log_{4}y = \dfrac{ 2 \log_{4}5}{\log_{6}5} $
$ \Rightarrow \log_{4}y = \dfrac{ 2 \log_{4}5}{\frac{1}{\log_{5}6}} \quad \left[ \because \log_{a}b = \frac{1}{\log_{b}a}\right] $
$ \Rightarrow \log_{4}y = 2(\log_{4}5)(\log_{5}6) $
$ \Rightarrow \log_{4}y = 2 \left( \frac{\log_{5}6}{\log_{5}4} \right) $
$ \Rightarrow \log_{4}y = 2 \log_{4}6 $
$ \Rightarrow \log_{4}y = \log_{4} 6^{2} $
$ \Rightarrow y = 6^{2} $
$ \Rightarrow \boxed{y = 36} $
Correct Answer$: 36$