Question

Arrange the following in ascending order of magnitude:

(i) $\sqrt[3]{4},\sqrt[4]{5},\sqrt{3}$

Answer 1

We have $\sqrt[3]{4},\sqrt[4]{5}\mathrm{and}\sqrt{3}$.
The orders of $\sqrt[3]{4},\sqrt[4]{5}\mathrm{and}\sqrt{3}$ are 3, 4 and 2.

The LCM of 3, 4 and 2 is 12.
So, we will make the order of all the surds 12.
$$\begin{gathered} \sqrt[3]{4}=\sqrt[3\times 4]{4^4}=\sqrt[12]{256} \\ \sqrt[4]{5}=\sqrt[4\times 3]{5^3}=\sqrt[12]{125} \\ \mathrm{and}\sqrt[]{3}=\sqrt[2\times 6]{3^6}=\sqrt[12]{729} \end{gathered}$$

Hence, the order is same.
Since 125< 256 < 729, $\sqrt[4]{5}<\sqrt[3]{4}<\sqrt{3}$.