Question

The value of ${\left(\right({\log }_{2}9{)}^2)}^{\frac{1}{{\log }_2\left({\log }_{2}9\right)}}\cdot {\left(\sqrt{7}\right)}^{\frac{1}{{\log }_{4}7}}$ is

Answer 1

${\left(\right({\log }_{2}9{)}^2)}^{\frac{1}{{\log }_2\left({\log }_{2}9\right)}}\cdot {\left(\sqrt{7}\right)}^{\frac{1}{{\log }_{4}7}}$
$={\left({\log }_{2}9\right)}^{2{\log }_{\left({\log }_{2}9\right)}2}\cdot {\left(\sqrt{7}\right)}^{{\log }_{7}4}$
$={\left({\log }_{2}9\right)}^{{\log }_{\left({\log }_{2}9\right)}4}\cdot {\left(4\right)}^{{\log }_7\sqrt{7}}$
$=4\cdot 4^{1/2}(\because x^{{\log }_{x}y}=y^{{\log }_{x}x}=y)$
$=4\cdot 2=8$