Arrange the following in ascending order of magnitude:
$\sqrt[3]{4}, \sqrt[4]{5}, \sqrt{3}$

\sqrt[4]{5} < \sqrt[3]{4} < \sqrt{3}

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\sqrt[4]{5} > \sqrt[3]{4} > \sqrt{3}

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\sqrt[4]{5} > \sqrt[3]{4} < \sqrt{3}

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\sqrt[4]{5} < \sqrt[3]{4} > \sqrt{3}

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Solution

The correct option is D $\sqrt[4]{5} < \sqrt[3]{4} < \sqrt{3}$
$\sqrt[3]{4}, \sqrt[4]{5} \sqrt{3}$
To compare, the order of the surds must be same.
$=> 4^{\frac{1}{3}}, 5^{\frac{1}{4}} 3^{\frac{1}{2}}$
$=> 4^{\frac{4}{12}}, 5^{\frac{3}{12}} 3^{\frac{6}{12}}$
$=> \sqrt[12]{4^{4}}, \sqrt[12]{5^{3}} \sqrt[12]{3^{6}}$
$=> \sqrt[12]{256}, \sqrt[12]{125} \sqrt[12]{729}$
Therefore,
$=> 125 < 256 < 729$
$=>, \sqrt[12]{125} < \sqrt[12]{256} < \sqrt[12]{729}$
$=> \sqrt[4]{5} < \sqrt[3]{4} < \sqrt{3}$

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