Question
Write the following in descending order:$\sqrt{6}, \sqrt[3]{8}$ and $\sqrt[4]{3}$
✓
Answer
Since $\sqrt{6}=6^{\frac{1}{2}}$ has power $\frac{1}{2}$,
$\sqrt[3]{8}=2$
$\sqrt[4]{3}=3^{\frac{1}{4}}$ has power $\frac{1}{4}$
Now,$\text{L.C.M}.$ of $2, 1$ and $4=4$
$\therefore \sqrt{6}=6^{\frac{1}{2}}=6^{\frac{2}{4}}=\left(6^2\right)^{\frac{1}{4}}=(36)^{\frac{1}{4}} $
$\sqrt[3]{8}=2=2^{\frac{4}{4}}=\left(2^4\right)^{\frac{1}{4}}=(16)^{\frac{1}{4}}$
$\sqrt[4]{3}=3^{\frac{1}{4}}=\left(3^1\right)^{\frac{1}{4}}=(3)^{\frac{1}{12}}$
Since, $36>16>3$,
we have $(36)^{\frac{1}{4}}>(16)^{\frac{1}{4}}>(3)^{\frac{1}{12}}$.
Hence, $\sqrt{6}>\sqrt[3]{8}>\sqrt[4]{3}$
$\sqrt[3]{8}=2$
$\sqrt[4]{3}=3^{\frac{1}{4}}$ has power $\frac{1}{4}$
Now,$\text{L.C.M}.$ of $2, 1$ and $4=4$
$\therefore \sqrt{6}=6^{\frac{1}{2}}=6^{\frac{2}{4}}=\left(6^2\right)^{\frac{1}{4}}=(36)^{\frac{1}{4}} $
$\sqrt[3]{8}=2=2^{\frac{4}{4}}=\left(2^4\right)^{\frac{1}{4}}=(16)^{\frac{1}{4}}$
$\sqrt[4]{3}=3^{\frac{1}{4}}=\left(3^1\right)^{\frac{1}{4}}=(3)^{\frac{1}{12}}$
Since, $36>16>3$,
we have $(36)^{\frac{1}{4}}>(16)^{\frac{1}{4}}>(3)^{\frac{1}{12}}$.
Hence, $\sqrt{6}>\sqrt[3]{8}>\sqrt[4]{3}$
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