$T(n) = T(n-2) + 2log n$ ---- (1)
$T(1) = 1$
Putting n = 3 in (1),
$T(3) = T(1) + 2log 3$
or $T(3) = 1 + 2 log 3$
Similarly we get,
$T(5) = 1 + 2log3 + 2 log5 = 1 + 2(log 3+ log 5)$
$T(7) = 1 + 2(log 3 + log 5) + 2log 7 =1+2(log 3 + log 5 + log 7)$
Therefore,
$T(n) = 1 + 2(log 3 + log 5 + log 7... + log n)$
$T(n) = 1 + 2(log 3.5.7..n)$ //log a+log b = log (a.b)
So, $T(n)< log n^n$ //check by putting values for n
which is $O(nlogn)$.