Question

Find the value of $\frac{{\log }_{5}9\cdot {\log }_{7}5\cdot {\log }_{3}7}{{\log }_2\sqrt{6}}+\frac{1}{{\log }_9\sqrt{6}}$.

Answer 1

Answer: 4

$\frac{{\log }_{5}9\cdot {\log }_{7}5\cdot {\log }_{3}7}{{\log }_2\sqrt{6}}+\frac{1}{{\log }_9\sqrt{6}}$
$=\frac{{\log }_{3}9}{{\log }_2\sqrt{6}}+\frac{1}{{\log }_9\sqrt{6}}=\frac{2}{{\log }_2\sqrt{6}}+\frac{1}{{\log }_9\sqrt{6}}=\frac{2}{\frac{1}{{\log }_{\sqrt{6}}2}}+\frac{\frac{1}{{\log }_{\sqrt{6}}9}}{}=2{\log }_{\sqrt{6}}2+{\log }_{\sqrt{6}}9$
$={\log }_{\sqrt{6}}4+{\log }_{\sqrt{6}}9={\log }_{\sqrt{6}}36=4$