The product sqrt[3]{2} cdot sqrt[4]{2} cdot s | vedclass

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

Question

(C) We have,$\sqrt[3]{2} \cdot \sqrt[4]{2} \cdot \sqrt[12]{32} = 2^{\frac{1}{3}} \cdot 2^{\frac{1}{4}} \cdot (2^5)^{\frac{1}{12}}$$= 2^{\frac{1}{3}} \

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) We have,$\sqrt[3]{2} \cdot \sqrt[4]{2} \cdot \sqrt[12]{32} = 2^{\frac{1}{3}} \cdot 2^{\frac{1}{4}} \cdot (2^5)^{\frac{1}{12}}$$= 2^{\frac{1}{3}} \cdot 2^{\frac{1}{4}} \cdot 2^{\frac{5}{12}}$Using the law of exponents $a^m \cdot a^n = a^{m+n}$,we get:$= 2^{\frac{1}{3} + \frac{1}{4} + \frac{5}{12}}$$= 2^{\frac{4+3+5}{12}}$$= 2^{\frac{12}{12}}$$= 2^1 = 2$Hence,$(c)$ is the correct answer.

The product $\sqrt[3]{2} \cdot \sqrt[4]{2} \cdot \sqrt[12]{32}$ equals

Similar Questions

The decimal expansion of the number $\sqrt{2}$ is

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

For each question,select the proper option from four options given,to make the statement true: The decimal expansion of $-\frac{11}{4}$ is..........

If $\sqrt{2} = 1.4142$,then evaluate $\sqrt{5} \div \sqrt{10}$ correct to four decimal places.

Find the value of $a$ in the following:
$\frac{6}{3 \sqrt{2}-2 \sqrt{3}}=3 \sqrt{2}-a \sqrt{3}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module