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Over the reals, an exponential equation answers ln(-1) as a root #2425
Description
Under MathS.Settings.Codomain.Set(Domain.Real), an exponential equation can answer a root that is not real. Measured on master bbf56b5:
| equation | gives | over the reals |
|---|---|---|
e^x + 1 = 0 |
{ ln(-1) } |
{ } |
e^x - 2 e^(-x) - 1 = 0 |
{ ln(-1), ln(2) } |
{ ln(2) } |
e^x - 2 = 0 |
{ ln(2) } |
{ ln(2) } |
ln(-1) is i pi, so it is not a real root. The second equation is t^2 - t - 2 = 0 in t = e^x, with roots t = 2 and t = -1. Only t = 2 has a real x.
Found while investigating #2319.
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