Question
Compare the following:$\sqrt[4]{12}$ and $\sqrt[3]{15}$
✓
Answer
$\sqrt[4]{12}=12^{\frac{1}{4}}$ has power $\frac{1}{4} $
$\sqrt[3]{15}=15^{\frac{1}{3}}$ has power $\frac{1}{3}$
Now, $\text{L.C.M}$. of $4$ and $3=12$
$\sqrt[4]{12}=12^{\frac{1}{4}}=12^{\frac{3}{12}}=\left(12^3\right)^{\frac{1}{12}}=(1728)^{\frac{1}{12}} $
$\sqrt[3]{15}=15^{\frac{1}{3}}=15^{\frac{4}{12}}=\left(15^4\right)^{\frac{1}{12}}=(50625)^{\frac{1}{12}}$
Since $1728<50625$,
we have $(1728)^{\frac{1}{12}}<(50625)^{\frac{1}{12}}$.
Hence, $\sqrt[4]{12}<\sqrt[3]{15}$.
$\sqrt[3]{15}=15^{\frac{1}{3}}$ has power $\frac{1}{3}$
Now, $\text{L.C.M}$. of $4$ and $3=12$
$\sqrt[4]{12}=12^{\frac{1}{4}}=12^{\frac{3}{12}}=\left(12^3\right)^{\frac{1}{12}}=(1728)^{\frac{1}{12}} $
$\sqrt[3]{15}=15^{\frac{1}{3}}=15^{\frac{4}{12}}=\left(15^4\right)^{\frac{1}{12}}=(50625)^{\frac{1}{12}}$
Since $1728<50625$,
we have $(1728)^{\frac{1}{12}}<(50625)^{\frac{1}{12}}$.
Hence, $\sqrt[4]{12}<\sqrt[3]{15}$.
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