VLEWhy

UsingExponential

It is less risky numerically to solve

exp(y-5) - x = 0

than to solve its equivalent

y-5 - ln(x)  = 0

The latter form will fail to evaluate if the solution for x wanders into the negative region during the solving process. Also, the derivative of ln(x) with respect to x (needed in evaluating the Jacobian for the Newton method) is 1/x, which heads to infinity as x tends to zero. So the Jacobian will "blow up" on you for small x.

The first form has none of these numerical problems.