sqrt[4]{sqrt[3]{2^{2}}} equals | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) To solve the expression $\sqrt[4]{\sqrt[3]{2^{2}}}$,we use the laws of exponents.First,express the radicals as fractional exponents: $\sqrt[n]{x}

Vedclass · Accepted Answer

$2^{\frac{1}{6}}$

Vedclass · Answer

(B) To solve the expression $\sqrt[4]{\sqrt[3]{2^{2}}}$,we use the laws of exponents.First,express the radicals as fractional exponents: $\sqrt[n]{x} = x^{\frac{1}{n}}$.$\sqrt[4]{\sqrt[3]{2^{2}}} = \sqrt[4]{(2^{2})^{\frac{1}{3}}}$.Using the power of a power rule $(a^{m})^{n} = a^{m 	imes n}$,we get $(2^{2})^{\frac{1}{3}} = 2^{\frac{2}{3}}$.Now,apply the outer radical: $(2^{\frac{2}{3}})^{\frac{1}{4}} = 2^{\frac{2}{3} 	imes \frac{1}{4}}$.Simplifying the exponent: $\frac{2}{3} 	imes \frac{1}{4} = \frac{2}{12} = \frac{1}{6}$.Thus,the expression simplifies to $2^{\frac{1}{6}}$.Hence,$(B)$ is the correct answer.

$\sqrt[4]{\sqrt[3]{2^{2}}}$ equals

Similar Questions

Insert a rational number and an irrational number between $2$ and $3$.

Find the value of the following expression correct to three decimal places,rationalizing the denominator if needed. Given $\sqrt{2} = 1.414$,$\sqrt{3} = 1.732$,and $\sqrt{5} = 2.236$.
$\frac{1}{\sqrt{3} + \sqrt{2}}$

Find the value of $\frac{4}{(216)^{-\frac{2}{3}}} + \frac{1}{(256)^{-\frac{3}{4}}} + \frac{2}{(243)^{-\frac{1}{5}}}$.

Insert a rational number and an irrational number between the following: $\frac{1}{3}$ and $\frac{1}{2}$

Show that $0.1\overline{6} = \frac{1}{6}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module