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Mathematics 22 Online
OpenStudy (anonymous):

\[-\int e^{t} dt = -e^{t}\] \[ln(-e^{t}) = -t?\]

OpenStudy (anonymous):

Over the real numbers \[\ln(-e^t)\] doesn't make any sense. Over the complex numbers however, this is given by: \[e^{i \pi} = -1 \implies \ln(-e^t)=\ln(e^{i \pi}e^t)=\ln(e^{i \pi})+\ln(e^t)=i \pi + t\]

OpenStudy (anonymous):

I.e, when dealing with real numbers you can't put negatives into logarithms.

OpenStudy (anonymous):

thanks

OpenStudy (anonymous):

@malevolence19 can you take a look at a new question that I have posted? That's where this question originated from

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