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lgbasallote

  • 3 years ago

Lgbariddle!! is \[\huge a^{\frac mn} = \sqrt[n]{a^m} = (\sqrt[n] a)^m\]

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  1. decripter37
    • 3 years ago
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    yup

  2. punnus
    • 3 years ago
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    Can be proved

  3. lgbasallote
    • 3 years ago
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    so tell me...what happens when a is negative?

  4. lgbasallote
    • 3 years ago
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    for example \[\huge (-1)^{2/4}\]

  5. lgbasallote
    • 3 years ago
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    \[\huge \sqrt[4]{-1} \implies \text{imaginary}\] \[\huge \sqrt[4] {(-1)^2} \implies 1\]

  6. decripter37
    • 3 years ago
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    you're right, again

  7. lgbasallote
    • 3 years ago
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    that's where your algebra is now shaken :troll:

  8. lgbasallote
    • 3 years ago
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    this is the part where i advertise this: http://openstudy.com/updates/4fdaecf3e4b0f2662fd13fd6

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