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Maths-

General

Easy

Question

Arrange in descending order of magnitude $cube root of 2 comma root index 6 of 3$ and $root index 9 of 4$

hint Hint:

Use LCM property.

The correct answer is: descending order

Complete step by step solution:
Here we can write, $root index 8 of 2 equals 2 to the power of 1 third end exponent comma space of 1em root index 6 of 3 equals 3 to the power of 1 over 6 end exponent$ and $root index 9 of 4 equals 4 to the power of 1 over 9 end exponent$
On taking the LCM of 3,6,9 we have LCM as 18.
So, $not stretchy rightwards double arrow 2 to the power of 1 third end exponent equals 2 to the power of fraction numerator 1 cross times 6 over denominator 3 cross times 6 end fraction end exponent equals 2 to the power of 6 over 18 end exponent$
$not stretchy rightwards double arrow open parentheses 2 to the power of 6 close parentheses to the power of 1 over 18 end exponent equals 64 to the power of 1 over 18 end exponent space of 1em open parentheses text Since end text open parentheses a to the power of b close parentheses to the power of c equals a to the power of b c end exponent close parentheses$
Likewise, $not stretchy rightwards double arrow 3 to the power of 1 over 6 end exponent equals 3 to the power of fraction numerator 1 cross times 3 over denominator 6 cross times 3 end fraction end exponent equals 3 to the power of 3 over 18 end exponent$
$not stretchy rightwards double arrow open parentheses 3 cubed close parentheses to the power of 1 over 18 end exponent equals 27 to the power of 1 over 18 end exponent space of 1em open parentheses text Since end text open parentheses a to the power of b close parentheses to the power of c equals a to the power of b c end exponent close parentheses$
Likewise, $not stretchy rightwards double arrow 4 to the power of 1 over 9 end exponent equals 4 to the power of fraction numerator 1 cross times 2 over denominator 9 cross times 2 end fraction end exponent equals 4 to the power of 2 over 18 end exponent$
$not stretchy rightwards double arrow open parentheses 4 squared close parentheses to the power of 1 over 18 end exponent equals 16 to the power of 1 over 18 end exponent space of 1em open parentheses text Since end text open parentheses a to the power of b close parentheses to the power of c equals a to the power of b c end exponent close parentheses$
Now, we have $64 to the power of 1 over 18 end exponent comma 27 to the power of 1 over 18 end exponent$ and $16 to the power of 1 over 18 end exponent$
So, here descending order is 64 > 27 > 16
$not stretchy rightwards double arrow 64 to the power of 1 over 18 end exponent greater than 27 to the power of 1 over 18 end exponent greater than 16 to the power of 1 over 18 end exponent$
$table attributes columnspacing 1em end attributes row cell not stretchy rightwards double arrow 2 to the power of 1 third end exponent greater than 3 to the power of 1 over 6 end exponent greater than 4 to the power of 1 over 9 end exponent end cell row cell not stretchy rightwards double arrow cube root of 2 greater than root index 6 of 3 greater than root index 9 of 4 end cell end table$

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