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The least number among $\sqrt[3]{4}, \sqrt[4]{5}, \sqrt[4]{7}$ and $\sqrt[3]{8}$
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Verified Answer
The correct answer is:
$\sqrt[4]{5}$
We have, $\sqrt[3]{4}, \sqrt[4]{5}, \sqrt[4]{7}, \sqrt[3]{8}$
$\begin{gathered}
\equiv(4)^{1 / 3},(5)^{1 / 4},(7)^{1 / 4},(8)^{1 / 3} \\
\equiv\left[(4)^{1 / 3}\right]^{12}, \\
{\left[(5)^{1 / 4}\right]^{12},\left[(7)^{1 / 4}\right]^{12},\left[(8)^{1 / 3}\right]^{12}} \\
=(4)^4,(5)^3,(7)^3,(8)^4
\end{gathered}$
Clearly, $(5)^3$ i.e., $\sqrt[4]{5}$ is least.
$\begin{gathered}
\equiv(4)^{1 / 3},(5)^{1 / 4},(7)^{1 / 4},(8)^{1 / 3} \\
\equiv\left[(4)^{1 / 3}\right]^{12}, \\
{\left[(5)^{1 / 4}\right]^{12},\left[(7)^{1 / 4}\right]^{12},\left[(8)^{1 / 3}\right]^{12}} \\
=(4)^4,(5)^3,(7)^3,(8)^4
\end{gathered}$
Clearly, $(5)^3$ i.e., $\sqrt[4]{5}$ is least.
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