When A = $\sqrt[3]{3}$ $root(3)3$ , B = $\sqrt[4]{4}$ $root(4)4$ , C = $\sqrt[6]{6}$ $root(6)6$ , find the relationship. which number is the correct number? 1. A<B<C 2. A<C<B 3. B<A<C 4. B<C<A 5. C<B<A

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Answer 1 · 2018-04-20T10:16:29

$5 . C < B < A$ $5 . C < B < A$

Here,

$A = \sqrt[3]{3}, B = \sqrt[4]{4} and C = \sqrt[6]{6}$ $A=root(3)3, B=root(4)4 and C=root(6)6$

Now, $LCM of : 3 , 4 , 6 is 12$ $"LCM of : 3 , 4 , 6 is 12"$

So,

$A^{12} = {(\sqrt[3]{3})}^{12} = {(3^{\frac{1}{3}})}^{12} = 3^{4} = 81$ $A^12=(root(3)3)^12=(3^(1/3))^12=3^4=81$

$B^{12} = {(\sqrt[4]{4})}^{12} = {(4^{\frac{1}{4}})}^{12} = 4^{3} = 64$ $B^12=(root(4)4)^12=(4^(1/4))^12=4^3=64$

$C^{12} = {(\sqrt[6]{6})}^{12} = {(6^{\frac{1}{6}})}^{12} = 6^{2} = 36$ $C^12=(root(6)6)^12=(6^(1/6))^12=6^2=36$

$i . e . 36 < 64 < 81$ $i.e. 36 < 64 < 81$

$\Rightarrow C^{12} < B^{12} < A^{12}$ $=>C^12 < B^12 < A^12$

$\Rightarrow C < B < A$ $=> C < B < A$

When A = 3√3root(3)3, B = 4√4root(4)4, C = 6√6root(6)6 , find the relationship. which number is the correct number? 1. A<B<C 2. A<C<B 3. B<A<C 4. B<C<A 5. C<B<A