Is that a number 1 in your exponential or a letter I? If it's the number 1 then the simple answer that you've been given will work. If it's the letter I then you have an algebraic equation of x=I+ln(I). Perhaps you could construct your array in reverse (i.e., determine x in terms of I) and then plot it the other way. I have attached a simple example. You could use a case structure in conjunction with a while loop instead of the for loop to set upper and lower limits on the value of x. Note that this will not give you equidistant x values.
There are certainly ways to solve for the value of I, but at present I can't think of any which are straightforward.
Unfortunately it is the letter I. Perhaps Y=exp(X-Y) is clearer. And also I need to find Y for a given X not the reverse. Just plotting it the other way is difficult since Y will be varying over orders of magnitude while X isn't. And I would like it to be more general.
I could do this with the full out Lev-Marq but that seemed way overkill. I'm still hoping for a simple solution with a zero finder or something.
