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Evaluate

$\frac{1}{n^{\log_{2}n}}$
asked in Numerical Ability by Veteran (11.2k points)   | 69 views

1 Answer

+3 votes
$log Y =log\left ( \frac{1}{n^{log_{2}n}} \right )$

$log Y =log1-log\left ( n^{log_{2}n} \right )$

$log Y =-\left (log_{2} n\right )^{2}$

$Y =2^{-\left (log_{2} n\right )^{2}}$
answered by Veteran (58.4k points)  
edited by

 

logY=log(1nlog2n)logY=log(1nlog2n)

logY=log1−log(nlog2n)logY=log1−log(nlog2n)

logY=−(log2n)2logY=−(log2n)2

Y=e−(log2n)2  // how is this line?

@srestha

it will be

Y = $2^{-(\log_{2}n)^{2}}$

you are using log(base 2) everywhere, then why at last e^ ....
yah, edited..


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