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Mathematics 12 Online
OpenStudy (anonymous):

e^ln2 ?

OpenStudy (anonymous):

2

OpenStudy (anonymous):

so e^-ln2 = -2 right?

OpenStudy (anonymous):

no...

OpenStudy (anonymous):

1/2?

OpenStudy (anonymous):

yes:)

OpenStudy (anonymous):

how?

OpenStudy (anonymous):

ln(a^b) = b*ln(a) So ln(1/2) = ln(2^(-1)) = - ln (2)

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