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Arrange the following numbers in the ascending order :
$\sqrt{3},\ \sqrt[3]{4},\ \sqrt[4]{10}$
Given: Numbers $\sqrt{3},\ \sqrt[3]{4},\ \sqrt[4]{10}$.
To do: To arrange the given numbers in ascending order.
Solution:
Given numbers:
$\sqrt{3}=3^{\frac{1}{2}}$
$\sqrt[3]{4}=4^{\frac{1}{3}}$
$\sqrt[4]{10}=10^{\frac{1}{4}}$
LCM of the denominators of the exponents $2,\ 3,\ 4=12$
Therefore, $\sqrt{3}=3^{\frac{1}{2}}$
$=3^{\frac{1}{2}\times\frac{6}{6}}$
$=3^{\frac{6}{12}}$
$=729^{\frac{1}{12}}$
$=\sqrt[12]{729}$
$\sqrt[3]{4}=4^{\frac{1}{3}}$
$=4^{\frac{1}{3}\times\frac{4}{4}}$
$=4^{\frac{4}{12}}$
$=( 4^4)^{\frac{1}{12}}$
$=256^{\frac{1}{12}}$
$=\sqrt[12]{256}$
And $\sqrt[4]{10}=10^{\frac{1}{4}}$
$=10^{\frac{1}{4}\times\frac{3}{3}}$
$=10^{\frac{3}{12}}$
$=( 10^3)^{\frac{1}{12}}$
$=1000^{\frac{1}{12}}$
$=\sqrt[12]{1000}$
On arranging the given numbers in ascending order
$\sqrt[12]{256}<\sqrt[12]{729}<\sqrt[12]{1000}$
Or
Updated on: 10-Oct-2022
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