Question

Simplify $\frac{{81}^{\frac{1}{{\log }_{5}9}}+3^{\frac{3}{{\log }_{\sqrt{6}}3}}}{409}\left({\left(\sqrt{7}\right)}^{\frac{2}{{\log }_{25}7}}-{\left(125\right)}^{{\log }_{25}6}\right)$

Answer 1

$\frac{{81}^{\frac{1}{{\log }_{5}9}}+3^{\frac{3}{{\log }_{\sqrt{6}}3}}}{409}\left(\right(\sqrt{7}{)}^{\frac{2}{{\log }_{25}7}}-\left(125{)}^{{\log }_{25}6}\right)$

$=\frac{9^{\left(2{\log }_{9}5\right)}+3^{\left(3{\log }_3\sqrt{6}\right)}}{409}\left(\right(7{)}^{1/2\times 2{\log }_{7}25}-\left(5{)}^{3/2{\log }_{5}6}\right)$

$=\frac{9^{{\log }_{9}25}+3^{{\log }_3{\left(6\right)}^{3/2}}}{409}\left(\right(7{)}^{{\log }_{7}25}-\left(5{)}^{{\log }_5{\left(6\right)}^{3/2}}\right)$

$=\frac{(25+(6{)}^{3/2})}{409}(25-(6{)}^{3/2})$

$=\frac{(25{)}^2-(6{)}^{3/2\times 2}}{409}$

$=\frac{625-216}{409}$

$=1$