The value of ${81}^{\left(1/{\log }_{5}3\right)}+{27}^{{\log }_{9}36}+3^{\left(4/{\log }_{7}9\right)}$ is

We know that $a^{{\log }_{a}b}=b$

${81}^{\left(1/{\log }_{5}3\right)}+{27}^{{\log }_{9}36}+3^{\left(4/{\log }_{7}9\right)}$
$=(3^4{)}^{{\log }_{3}5}+(3^3{)}^{{\log }_{3^2}{6}^2}+3^{4{\log }_{9}7}$
$=3^{{\log }_3{5}^4}+(3^3{)}^{{\log }_{3}6}+3^{4{\log }_{3^2}7}$
$=5^4+3^{{\log }_3{6}^3}+3^{2{\log }_{3}7}$
$=625+6^3+3^{{\log }_3{7}^2}$
$=625+216+7^2$
$=890$