To simplify a bit I'll assume that N and M are > k, so the complexity here is O(log k), which is O(log N + log M).
Pseudo-code:
i = k/2
j = k - i
step = k/4
while step > 0
if a[i-1] > b[j-1]
i -= step
j += step
else
i += step
j -= step
step /= 2
if a[i-1] > b[j-1]
return a[i-1]
else
return b[j-1]