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<p><b>New page</b></p><div>===UsingExponential===<br />
It is less risky numerically to solve <br />
<br />
<source lang="a4c">exp(y-5) - x = 0</source><br />
<br />
than to solve its equivalent <br />
<br />
<source lang="a4c">y-5 - ln(x) = 0</source><br />
<br />
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.<br />
<br />
The first form has none of these numerical problems.<br />
<br />
<br />
<br />
<br />
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