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

Evaluate: $\frac{2 \cdot 3^{n+1} + 7 \cdot 3^{n-1}}{3^{n+2} - 2 \cdot (1/3)^{1-n}}$

If $x = 3 - \sqrt{5}$,then find the value of $\frac{\sqrt{x}}{\sqrt{2} + \sqrt{3x - 2}}$.

If $\frac{4 + 3\sqrt{3}}{\sqrt{7 + 4\sqrt{3}}} = a + \sqrt{b}$,then $(a, b) = $

$\frac{12}{3 + \sqrt{5} - 2\sqrt{2}} = $

$\frac{\sqrt{5/2} + \sqrt{7 - 3\sqrt{5}}}{\sqrt{7/2} + \sqrt{16 - 5\sqrt{7}}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module