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OpenStudy (anonymous):
Evaluate e^ln1
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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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