Write the following in ascending order of magnitude.
Question:

Write the following in ascending order of magnitude.

$\sqrt[6]{6}, \sqrt[3]{7}, \sqrt[4]{8}$

 

Solution:

$\sqrt[6]{6}, \sqrt[3]{7}, \sqrt[4]{8}$

$\sqrt[6]{6}=(6)^{\frac{1}{6}}=(6)^{\frac{2}{12}}=\left(6^{2}\right)^{\frac{1}{12}}=(36)^{\frac{1}{12}} \quad \ldots(1)$

$\sqrt[3]{7}=(7)^{\frac{1}{3}}=(7)^{\frac{4}{12}}=\left(7^{4}\right)^{\frac{1}{12}}=(2401)^{\frac{1}{12}} \ldots(2)$

$\sqrt[4]{8}=(8)^{\frac{1}{4}}=(8)^{\frac{3}{12}}=\left(8^{3}\right)^{\frac{1}{12}}=(512)^{\frac{1}{12}} \ldots(3)$

On Comparing $(1),(2)$ and $(3)$, we get

$(36)^{\frac{1}{12}}<(512)^{\frac{1}{12}}<(2401)^{\frac{1}{12}}$

$\Rightarrow \sqrt[6]{6}<\sqrt[4]{8}<\sqrt[3]{7}$

Hence, $\sqrt[6]{6}<\sqrt[4]{8}<\sqrt[3]{7}$.

 

 

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