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spndsh

  • 3 years ago

e^ln2 ?

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  1. rsilvabenevides
    • 3 years ago
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    2

  2. spndsh
    • 3 years ago
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    so e^-ln2 = -2 right?

  3. Algebraic!
    • 3 years ago
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    no...

  4. spndsh
    • 3 years ago
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    1/2?

  5. Algebraic!
    • 3 years ago
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    yes:)

  6. spndsh
    • 3 years ago
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    how?

  7. rsilvabenevides
    • 3 years ago
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    ln(a^b) = b*ln(a) So ln(1/2) = ln(2^(-1)) = - ln (2)

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