Question

The ascending order of the surds $\sqrt[3]{32},\sqrt[6]{63},\sqrt[9]{94}$ is

• A. $\sqrt[9]{94},\sqrt[6]{63},\sqrt[3]{32}$
• B. $\sqrt[9]{94},\sqrt[3]{32},\sqrt[6]{63}$
• C. $\sqrt[3]{32},\sqrt[6]{63},\sqrt[9]{94}$
• D. $\sqrt[6]{63},\sqrt[9]{94},\sqrt[3]{32}$

Answer 1

The correct option is A $\sqrt[9]{94},\sqrt[6]{63},\sqrt[3]{32}$

Lets keep in mind the following common log values:

$log\left(2\right)=0.30$

$log\left(3\right)=0.47$

Lets compare these three surds by their log values.

Reason: Since $log\left(x\right)$ is an increasing function when $x>1$ so we can compare these values by comparing their log values.

$log\left(\sqrt[3]{32}\right)=log\left(2^{\frac{1}{3}}\right)=\left(\frac{1}{3}\right)log\left(2\right)=\left(\frac{1}{3}\right)\times 0.3=0.100$

$log\left(\sqrt[6]{63}\right)=log\left(3^{\frac{1}{6}}\right)=\left(\frac{1}{6}\right)log\left(3\right)=\left(\frac{1}{6}\right)\times 0.47=0.078$

$log\left(\sqrt[9]{94}\right)=log\left(\sqrt[9]{92^2}\right)=log\left(2^{\frac{2}{9}}\right)=\left(\frac{2}{9}\right)log\left(2\right)=\left(\frac{2}{9}\right)\times 0.3=0.060$

By looking at the log values,it is clear that,

$\sqrt[9]{94}<\sqrt[6]{63}<\sqrt[3]{32}$