Which is the greatest number among sqrt[3]{9} | vedclass

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

Question

(A) To compare the numbers $\sqrt[3]{9}, \sqrt[4]{11}, \sqrt[6]{17}$,we convert them to the same root index.The least common multiple $(L.C.M.)$ of th

Vedclass · Accepted Answer

$\sqrt[3]{9}$

Vedclass · Answer

(A) To compare the numbers $\sqrt[3]{9}, \sqrt[4]{11}, \sqrt[6]{17}$,we convert them to the same root index.The least common multiple $(L.C.M.)$ of the indices $3, 4, 6$ is $12$.$\sqrt[3]{9} = 9^{1/3} = (9^4)^{1/12} = (6561)^{1/12}$$\sqrt[4]{11} = 11^{1/4} = (11^3)^{1/12} = (1331)^{1/12}$$\sqrt[6]{17} = 17^{1/6} = (17^2)^{1/12} = (289)^{1/12}$Comparing the values inside the brackets: $6561 > 1331 > 289$.Therefore,$\sqrt[3]{9}$ is the greatest number.

Which is the greatest number among $\sqrt[3]{9}, \sqrt[4]{11}, \sqrt[6]{17}$?

Similar Questions

$20^{2-3x^2} = (40\sqrt{5})^{3x^2-2}$,then $x$ is equal to

The number of solutions of the equation $(x)^{x\sqrt{x}} = (x\sqrt{x})^x$ is:

If $3^x - 3^{x - 1} = 6$,then $x^x$ is equal to

If $(a^m)^n = a^{m^n}$,then the value of $m$ in terms of $n$ is

Evaluate: $\sqrt{3 + \sqrt{5}}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module