Question 55

$4^{9}$ is greater than $16^{3} .$

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Solution

True
$∵ 16^{3} = (4^{2})^{3} = 4^{6}$
Now, in $4^{9} a n d 4^{9} > 4^{6}$ as powers 9>6 [if base is same, then power is greater number is also greater]

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