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Mathematics 19 Online
OpenStudy (anonymous):

Evaluate e^ln1

OpenStudy (anonymous):

against what?, if i may ask?

OpenStudy (anonymous):

\[e^{\ln a} = a\]

OpenStudy (anonymous):

what do you think is \(e^{\ln 1}\) then?

OpenStudy (anonymous):

another possible solution is \(\ln 1 = 0\) therefore \(e^0 = ?\)

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