The greatest of $$\sqrt 2 ,$$  $$\root 6 \of 3 ,$$  $$\root 3 \of 4 ,$$  $$\root 4 \of 5 $$   is = ?

Solution(By Examveda Team)

LCM of 2, 3, 4, 6 is 12
$$\sqrt 2 = $$  $${2^{\frac{1}{2}}} = $$  $${2^{\left( {\frac{1}{2} \times \frac{6}{6}} \right)}} = $$   $${2^{\frac{6}{{12}}}} = $$  $${\left( {{2^6}} \right)^{\frac{1}{{12}}}} = $$  $${\left( {64} \right)^{\frac{1}{{12}}}} = $$   $$\root {12} \of {64} $$
$$\root 6 \of 3 = $$  $${3^{\frac{1}{6}}} = $$  $${3^{\left( {\frac{1}{6} \times \frac{2}{2}} \right)}} = $$   $${3^{\frac{2}{{12}}}} = $$  $${\left( {{3^2}} \right)^{\frac{1}{{12}}}} = $$  $${\left( 9 \right)^{\frac{1}{{12}}}} = $$   $$\root {12} \of 9 $$
$$\root 3 \of 4 = $$  $${4^{\frac{1}{3}}} = $$  $${4^{\left( {\frac{1}{3} \times \frac{4}{4}} \right)}} = $$  $${4^{\frac{4}{{12}}}} = $$  $${\left( {{4^4}} \right)^{\frac{1}{{12}}}} = $$  $${\left( {256} \right)^{\frac{1}{{12}}}} = $$   $$\root {12} \of {256} $$
$$\root 4 \of 5 = $$  $${5^{\frac{1}{4}}} = $$  $${5^{\left( {\frac{1}{4} \times \frac{3}{3}} \right)}} = $$   $${5^{\frac{3}{{12}}}} = $$  $${\left( {{5^3}} \right)^{\frac{1}{{12}}}} = $$  $${\left( {125} \right)^{\frac{1}{{12}}}} = $$   $$\root {12} \of {125} $$
Clearly $$\root {12} \of {256} \,\,i.e.,\root 3 \of 4 $$     is the greatest.