Question
Write the following in descending order:$\sqrt{2}, \sqrt[3]{5}$ and $\sqrt[4]{10}$
✓
Answer
Since $\sqrt{2}=2^{\frac{1}{2}}$ has power $\frac{1}{2}$,
$\sqrt[3]{5}=5^{\frac{1}{3}}$ has power $\frac{1}{3}$
$\sqrt[4]{10}=10^{\frac{1}{4}}$ has power $\frac{1}{4}$
Now,$ \text{L.C.M.}$ of $2,3$ and $4=12$
$\therefore \sqrt{2}=2^{\frac{1}{2}}=2^{\frac{6}{12}}=\left(2^6\right)^{\frac{1}{12}}=(64)^{\frac{1}{12}}$
$\sqrt[3]{5}=5^{\frac{1}{2}}=5^{\frac{4}{12}}=\left(5^4\right)^{\frac{1}{12}}=(625)^{\frac{1}{12}}$
$\sqrt[4]{10}=10^{\frac{1}{4}}=10^{\frac{3}{12}}=\left(10^3\right)^{\frac{1}{12}}=(1000)^{\frac{1}{12}}$
Since, $1000>625>64$,
we have $(1000)^{\frac{1}{12}}>(625)^{\frac{1}{12}}>(64)^{\frac{1}{12}}$.
Hence, $\sqrt[4]{10}>\sqrt[3]{5}>\sqrt{2}$.
$\sqrt[3]{5}=5^{\frac{1}{3}}$ has power $\frac{1}{3}$
$\sqrt[4]{10}=10^{\frac{1}{4}}$ has power $\frac{1}{4}$
Now,$ \text{L.C.M.}$ of $2,3$ and $4=12$
$\therefore \sqrt{2}=2^{\frac{1}{2}}=2^{\frac{6}{12}}=\left(2^6\right)^{\frac{1}{12}}=(64)^{\frac{1}{12}}$
$\sqrt[3]{5}=5^{\frac{1}{2}}=5^{\frac{4}{12}}=\left(5^4\right)^{\frac{1}{12}}=(625)^{\frac{1}{12}}$
$\sqrt[4]{10}=10^{\frac{1}{4}}=10^{\frac{3}{12}}=\left(10^3\right)^{\frac{1}{12}}=(1000)^{\frac{1}{12}}$
Since, $1000>625>64$,
we have $(1000)^{\frac{1}{12}}>(625)^{\frac{1}{12}}>(64)^{\frac{1}{12}}$.
Hence, $\sqrt[4]{10}>\sqrt[3]{5}>\sqrt{2}$.
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