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anonymous

  • one year ago

Need Help! How do you solve for (3/7)^7/3 to be in fraction form?

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  1. mathmate
    • one year ago
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    Is the question \(\Large \frac{(\frac{3}{7})^7}{3}\) or is it \(\Large (\frac{3}{7})^\frac{7}{3}\) ? In any case, use \((\frac{A}{B})^X=\frac{A^X}{B^X}\) and \(M^\frac{x}{y}=\sqrt[y]{M^x}\) to help you.

  2. anonymous
    • one year ago
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    It's the second one!

  3. anonymous
    • one year ago
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    So if I were using the first method, it would be something like: 3^7/3 all over 7^7/3 ?

  4. anonymous
    • one year ago
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    But what would I do after that?

  5. mathmate
    • one year ago
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    @mary_am Please remember the priority of operations require that exponentiation is before multiplication and division, so use brackets as required. \(\Large (\frac{3}{7})^\frac{7}{3}\) is written (3/7)^(7/3) to avoid further confusion.

  6. mathmate
    • one year ago
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    "So if I were using the first method, it would be something like: 3^7/3 all over 7^7/3 ? " probably meant = ( 3^(7/3) ) / ( 7^(7/3) ) Yes written that way would be correct.

  7. anonymous
    • one year ago
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    Okay, and then what would happen?

  8. mathmate
    • one year ago
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    Then you would use the second rule to make it a simple fraction.

  9. anonymous
    • one year ago
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    |dw:1438826071903:dw|

  10. anonymous
    • one year ago
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    What would be the Z and the M?

  11. mathmate
    • one year ago
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    But that's a completely different problem! lol

  12. anonymous
    • one year ago
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    Oh, I was just going to figure out how to find (3/7)^(7/3) in its fraction form and try to apply the same method for the other value

  13. anonymous
    • one year ago
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    But I guess now that the air is clear, do you mind helping me with the problem I drew? lol

  14. mathmate
    • one year ago
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    I don't understand the second problem. \(\frac{3}{7}+\frac{4}{9}=\frac{55}{63}\) but not with the exponents that you included!

  15. anonymous
    • one year ago
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    Thats really weird... btw this came from me trying to solve an integral problem

  16. mathmate
    • one year ago
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    You may post the integral problem to start! :)

  17. anonymous
    • one year ago
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    |dw:1438826425496:dw|

  18. anonymous
    • one year ago
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    |dw:1438826526116:dw|

  19. anonymous
    • one year ago
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    .... and then I dont know what to do

  20. mathmate
    • one year ago
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    |dw:1438827006677:dw|

  21. anonymous
    • one year ago
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    Im trying to understand what you did at the bottom

  22. anonymous
    • one year ago
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    I learned to evaluate the integral by plugging in the top bound and bottom bound in the x values and then subtracting the top and bottom bound

  23. mathmate
    • one year ago
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    Yes, but it's the same as evaluating term by term, \([(3/7)*x^{7/3}+(4/9)x^{9/4}]_0^1=((3/7)*1^{7/3}+(4/9)1^{9/4})-((3/7)*^{7/3}+(4/9)0^{9/4}])\) \(=(3/7)+(4/9)-(0+0)=3/7+4/9=55/63\)

  24. mathmate
    • one year ago
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    note that 1^(7/3)=1, and 1^(9/4)=1, 0^(7/3)=0, and 0^(9/4)=0

  25. anonymous
    • one year ago
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    OH

  26. anonymous
    • one year ago
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    that makes so much sense, thanks!

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