The inequality asks for \(n\) where \(n^2-7n+14\) lies strictly between \(0\) and \(1\), because a base smaller than \(1\) flips the usual direction of a logarithmic inequality. Checking whether the quadratic can ever fall below \(1\) means checking whether \(n^2-7n+13\) (which is \(n^2-7n+14\) shifted down by \(1\)) can ever be negative. Its discriminant is \(49-4(13)=-3\), which is negative, so \(n^2-7n+13\) never touches zero and, since its leading coefficient is positive, stays positive for every real \(n\). That directly means \(n^2-7n+14\) never dips below \(1\), so it can never land in the required window \((0,1)\). With the target range completely unreachable, no value of \(n\), non-negative or otherwise, satisfies the inequality, giving \(\boxed{0}\) as the count.