If $x^{20} < 32^{18}$ , then the greatest possible integral value of 'x' is

22

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23

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24

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none of these

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Solution

The correct option is A $22$
$x^{20} < 32^{18} = x^{20} < 2^{90} (∵ 32 = 2^{5})$
$= {(x^{1 / 9})}^{180} < {(2^{1 / 2})}^{180}$
$= x^{1 / 9} < \sqrt{2} \Rightarrow x < 2^{9 / 2}$
$= x < 22.6$
$∴ x = 22$
is option $A$ is correct.

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