Question

The smallest of $\sqrt[3]{34},\sqrt[4]{45},\sqrt[4]{46},\sqrt[3]{38}$ is:

• A. $\sqrt[3]{38}$
• B. $\sqrt[4]{45}$
• C. $\sqrt[3]{34}$
• D. $\sqrt[4]{46}$

Answer 1

The correct option is B $\sqrt[4]{45}$

$\sqrt[3]{34}=\sqrt[12]{1244}=\sqrt[12]{12256}$

$\sqrt[4]{46}=\sqrt[12]{125^3}=\sqrt[12]{12125}$

$\sqrt[4]{46}=\sqrt[12]{126^3}=\sqrt[12]{12216}$

$\sqrt[3]{38}=\sqrt[12]{128^4}=\sqrt[12]{12{64}^2}$

As ${}^{\prime }{125}^{\prime }$ is smallest

$\therefore 4\sqrt{5}$ is smallest