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anonymous

  • one year ago

...Easy Medal!! Simplify 4 to the 7th over 5 squared all raised to the 3rd power. 4 to the 10th over 5 to the 5th 4 to the 4th over 5 4 to the 21st over 5 to the 6th 12 to the 7th over 15 squared

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  1. anonymous
    • one year ago
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    So is it like this? \[\large \sf \huge ( \large \frac{4^{7}}{5^{2}} \huge ) ^{3}\]

  2. anonymous
    • one year ago
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    yes

  3. anonymous
    • one year ago
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    @LegendarySadist

  4. anonymous
    • one year ago
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    Then you would use the exponent rule: \[\large \sf (b^{n})^m~=~b^{n \times m}\]

  5. anonymous
    • one year ago
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    which answer would that be

  6. anonymous
    • one year ago
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    \[\large \sf (\frac{b^{n}}{a^{l}})^m~=~\frac{b^{n \times m}}{a^{l \times m}}\]

  7. anonymous
    • one year ago
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    I don't get it

  8. anonymous
    • one year ago
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    That's not good

  9. anonymous
    • one year ago
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    ..

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spraguer (Moderator)
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