Question

Find the value of
${\log }_3{\log }_2{\log }_{\sqrt{3}}81$

Answer 1

${\log }_3{\log }_2{\log }_{\sqrt{3}}81={\log }_3{\log }_2{\log }_{\sqrt{3}}{3}^4={\log }_3{\log }_2{\log }_{\sqrt{3}}(\sqrt{3}{)}^8={\log }_3{\log }_{2}8={\log }_3{\log }_2{2}^3={\log }_{3}3=1$