Arranging the following in descending order,w | vedclass

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

Question

(A) To compare the given numbers,we express them with a common index (root).The numbers are $4^{1/3}, 2^{1/2}, 3^{1/6}, 5^{1/4}$.The denominators of t

Vedclass · Accepted Answer

$\sqrt[3]{4} > \sqrt[4]{5} > \sqrt{2} > \sqrt[6]{3}$

Vedclass · Answer

(A) To compare the given numbers,we express them with a common index (root).The numbers are $4^{1/3}, 2^{1/2}, 3^{1/6}, 5^{1/4}$.The denominators of the exponents are $3, 2, 6, 4$. The $LCM$ of these denominators is $12$.Now,rewrite each number with the exponent $1/12$:$4^{1/3} = 4^{4/12} = (4^4)^{1/12} = 256^{1/12}$$2^{1/2} = 2^{6/12} = (2^6)^{1/12} = 64^{1/12}$$3^{1/6} = 3^{2/12} = (3^2)^{1/12} = 9^{1/12}$$5^{1/4} = 5^{3/12} = (5^3)^{1/12} = 125^{1/12}$Comparing the bases: $256 > 125 > 64 > 9$.Therefore,the descending order is $\sqrt[3]{4} > \sqrt[4]{5} > \sqrt{2} > \sqrt[6]{3}$.

Arranging the following in descending order,we get $\sqrt[3]{4}, \sqrt{2}, \sqrt[6]{3}, \sqrt[4]{5}$.

Similar Questions

If $m$ and $n$ are positive integers and $(m-n)$ is an even number,then $(m^{2}-n^{2})$ will be always divisible by

$A$ fraction becomes $4$ when $1$ is added to both the numerator and denominator; and it becomes $7$ when $1$ is subtracted from both the numerator and denominator. The numerator of the given fraction is

$69.69 - 51.54 + 73.64 = ? + 32.42$

The units digit in $3 \times 38 \times 537 \times 1256$ is

If the sum and difference of two numbers are $20$ and $8$ respectively,then the difference of their squares is

Vedclass Test Series

Exam Paper Generator

Online Exam Module