Value of sqrt[4]{(81)^{-2}} is | vedclass

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

Question

(D) Given expression: $\sqrt[4]{(81)^{-2}}$We know that $a^{-n} = \frac{1}{a^n}$,so $(81)^{-2} = \frac{1}{81^2}$.Thus,$\sqrt[4]{(81)^{-2}} = \sqrt[4]{

Vedclass · Accepted Answer

$\frac{1}{9}$

Vedclass · Answer

(D) Given expression: $\sqrt[4]{(81)^{-2}}$We know that $a^{-n} = \frac{1}{a^n}$,so $(81)^{-2} = \frac{1}{81^2}$.Thus,$\sqrt[4]{(81)^{-2}} = \sqrt[4]{\frac{1}{81^2}} = \left(\frac{1}{81^2}ight)^{1/4}$.Since $81 = 9^2$,we have $81^2 = (9^2)^2 = 9^4$.Substituting this back,we get $\left(\frac{1}{9^4}ight)^{1/4} = \left(\frac{1}{9}ight)^{4 	imes \frac{1}{4}} = \frac{1}{9}$.Hence,$(d)$ is the correct answer.

Value of $\sqrt[4]{(81)^{-2}}$ is

Similar Questions

Simplify:
$\frac{9^{\frac{1}{3}} \times 27^{-\frac{1}{2}}}{3^{\frac{1}{6}} \times 3^{-\frac{2}{3}}}$

State whether the following statement is true or false:
Every whole number is a rational number.

Express $0.7\overline{39}$ in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$.

Insert a rational number and an irrational number between the following:
$6.375289$ and $6.375738$

Convert the following rational number into decimal form and state the kind of its decimal expansion:
$\frac{4}{13}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module