Here 1/2 is cancel out by 1 - 1/2
$n[1 - (1/2) ^\ log n]$
log property :: 2 ^ log n (base 2)
then replace n with 2
= n ^ log 2 = n
So, our equestion are in form ->n[1 - ( 2^(-1) ) ^logn]
so we can write as n[1 - (n^-1) ^ log2 ] then
= $n[ 1- 1/n ]$
= $n[ \frac{n-1}{n} ]$
= $ n - 1 $