log(x)

Probably because most other languages follow that same naming convention. Not sure why, just historical I guess.

Refer to the MATLAB "help" for these kind of questions...

The log function does exactly what you want. log(14 - y) If you want a base 10 log, you use log10. There is also a log2 function, which gives a base 2 log. Other bases are achieved using the simple relation log(X)/log(b) produces a log to the base b. You could write a function for it as: logb = @(X,b) log(X)/log(b); logb(9,3) ans = 2 which is as expected. Finally, there is the reallog function, which also does the natural log, but it produces an error when the log would have been a complex number. log(-2) ans = 0.693147180559945 + 3.14159265358979i reallog(-2) Error using reallog Reallog produced complex result. But for the more normal case, reallog does the same thing as log. log(2) ans = 0.693147180559945 reallog(2) ans = 0.693147180559945

Does a solution near(er) to zero exist? PLOT IT!!!!!! syms x f = (-27109/x)-0.95*log(x)+25.18 - log(0.5); fplot(f,[10,1000]) Do you see anything strange? A solution would correspond to a point where that curve crosses the line y == 0. Looking a little further up, we finally see this: fplot(f,[200,1e4]) yline(0); So a solution seems to occur near x == 1500. S = vpasolve(f,x,1500) However, it you look further out, the curve reaches a peak, and then starts to drop again. It may well cross zero again out there. fplot(f,[10000,1e6]) yline(0); Did you tell vpasolve which root it should find? Should it know? vpasolve is a numerical root finder. It looks for a root, near where you tell it to look. If you don't tell it where, it picks a spot. And sometimes, it may end up converging to a solution you don't like.

Did you try it with the proper parentheses? Sg2 = log(1-(Cdo-u*Clmin) / ((T/W - u) * Clrot)) What values are you passing in? What do you expect as a result and what are you getting? Answer after reading the posting guidelines: TUTORIAL: How to ask a question (on Answers) and get a fast answer