When I run your code (after fixing a typo when you refer to what I think should be b(n)/13), the result is a 1-by-6 array for c:
a =[27.7847; 31.1386; 33.3644; 37.0654; 38.5043; 41.3808];
b = [13; 33; 55; 155; 245; 519];
for n = 1:6
c(n)= a(n)*0.4/(log(b(n)/13));
end
c =
Inf 13.3705 9.2526 5.9820 5.2453 4.4894
To get c as a 6-by-1 vector instead, use c(n,1) inside the loop, or calculate the transpose c=c' after the loop is done.
Or even better, avoid the for loop altogether with the following vectorized assignment:
c = a.*0.4./(log(b./13))
This evaluates the expression one element at a time for a and b and constructs the vector c from the individual results.
