Which is greater √2,³√4,⁴√3
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Posted by Mamoni Saha 8 years, 6 months ago
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Rashmi Bajpayee 8 years, 6 months ago
{tex}\sqrt 2 ,\root 3 \of 4 ,\root 4 \of 3 {/tex}
= {tex}{2^{{1 \over 2}}},{4^{{1 \over 3}}},{3^{{1 \over 4}}}{/tex}
= {tex}{2^{{6 \over {12}}}},{4^{{4 \over {12}}}},{3^{{3 \over {12}}}}{/tex} [Making all the powers as like terms]
= {tex}\root {12} \of {{2^6}} ,\root {12} \of {{4^4}} ,\root {12} \of {{3^3}} {/tex}
= {tex}\root {12} \of {64} ,\root {12} \of {256} ,\root {12} \of {27} {/tex}
Here {tex}\root {12} \of {256} {/tex} is greater.
Therefore, {tex}\root 3 \of 4 {/tex} is greater.
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