Question

Find the value and don't spam and answer correctly ​

Answer 1

QUESTION:

$\sf\sqrt[5]{\sqrt[4]{3^2}}$

ANSWER:

$\sf\sqrt[5]{\sqrt[4]{3^2}}=3^{1\!/\!10}$

GIVEN:

$\sf\sqrt[5]{\sqrt[4]{3^2}}$

TO FIND:

$\sf\sqrt[5]{\sqrt[4]{3^2}}$

EXPLANATION:

We know that 3² = 9

$\sf\leadsto\sqrt[5]{\sqrt[4]{3^2}}$

9 = 3 × 3

3 = √3 × √3

9 = √3 × √3 × √3 × √3

$\sf\leadsto\sqrt[5]{\sqrt[4]{\sqrt{3}\times\sqrt{3}\times\sqrt{3}\times\sqrt{3}}}$

$\boxed{\bold{\large{\gray{If\ x^4= y\times y\times y\times y, \ x = y}}}}$

$\sf\leadsto\sqrt[5]{\sqrt{3}}$

$\boxed{\bold{\large{\gray{\sqrt{x} = x^{1\!/\!2}}}}}$

$\sf\leadsto\sqrt[5]{3^{1\!/\!2}}$

$\boxed{\bold{\large{\gray{\sqrt[5]{x} = x^{1\!/\!5}}}}}$

$\sf\leadsto (3^{1\!/\!2})^{1\!/\!5}$

$\boxed{\bold{\large{\gray{(x^m)^n = x^{mn}}}}}$

$\sf\leadsto 3^{1\!/\!10}$

$\sf\sqrt[5]{\sqrt[4]{3^2}}=3^{1\!/\!10}$

Hence the answer is 3^(1/10).

SOME MORE FORMULAE:

$\boxed{\bold{\large{\gray{x^m \times x^n = x^{m+n}}}}}$

$\boxed{\bold{\large{\gray{\dfrac{x^m}{x^n }= x^{m-n}}}}}$

$\boxed{\bold{\large{\gray{\dfrac{1}{x^n }= x^{-n}}}}}$

$\boxed{\bold{\large{\gray{\dfrac{1}{x^{-n} }= x^{n}}}}}$

SOME IDENTITIES:

$\boxed{\bold{\large{\gray{(A + B)^2 = A^2 + 2AB + B^2}}}}$

$\boxed{\bold{\large{\gray{(A - B)^2 = A^2 - 2AB + B^2}}}}$

$\boxed{\bold{\large{\gray{(A - B)(A+B) = A^2 - B^2}}}}$

$\boxed{\bold{\large{\gray{A^2 + B^2=(A+B)² - 2AB}}}}$

$\boxed{\bold{\large{\gray{A^2 + B^2=(A-B)² + 2AB}}}}$

Answer 2

Answer:

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