The greatest number among root 3 of 9 ,root 4 | vedclass

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

Question

(a) $oot 3 \of 9 ,\,\,oot 4 \of {11} ,\,\,oot 6 \of {17} $

$L.C.M.$ of $3, 4, 6$ is $12$

$	herefore \sqrt[3]{9} = {9^{1/3}} = {({9^4})^{1

Vedclass · Accepted Answer

$\root 3 \of 9 $

Vedclass · Answer

(a) $oot 3 \of 9 ,\,\,oot 4 \of {11} ,\,\,oot 6 \of {17} $

$L.C.M.$ of $3, 4, 6$ is $12$

$	herefore \sqrt[3]{9} = {9^{1/3}} = {({9^4})^{1/12}} = {(6561)^{1/12}}$

$oot 4 \of {11} = {(11)^{1/4}}{({11^3})^{1/12}} = {(1331)^{1/12}}$,

$oot 6 \of {17} = {(17)^{1/6}} = {({17^2})^{1/2}} = {(289)^{1/12}}$

Hence, $oot 3 \of 9 $ is the greatest number.

The greatest number among $\root 3 \of 9 ,\root 4 \of {11} ,\root 6 \of {17} $ is

Similar Questions

Let ${7 \over {{2^{1/2}} + {2^{1/4}} + 1}}$$ = A + B{.2^{1/4}} + C{.2^{1/2}} + D{.2^{3/4}}$, then $A+B+C+D= . . .$

Number of value/s of $x$ satisfy given eqution ${5^{x - 1}} + 5.{(0.2)^{x - 2}} = 26$.

The square root of $\frac{(0.75)^3}{1-(0.75)}+\left[0.75+(0.75)^2+1\right]$ is

The value of $\sqrt {[12\sqrt 5 + 2\sqrt {(55)} ]} $ is

The value of ${{15} \over {\sqrt {10} + \sqrt {20} + \sqrt {40} - \sqrt 5 - \sqrt {80} }}$ is