Originally Posted by

**Chrisgarrard** If y = ax^n then on a log x; log y plot (Eg in Excel) the gradient of the resulting straight line is n as logy = loga + nlogX. But in our particular problem y=a(x-b)^n (where b is a constant) and we do not know how to work out the new gradient on a logx ; logy plot despite many attempts with differential methods. By plotting examples in Excel it looks like the gradient has been modified to nx/(x-b). We would really like to show the working to prove this or otherwise. As we would like to include this solution in a publication we would happily acknowledge the author / source.

Thanks in advance,

Chris