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BecomeMyFan=D

  • 3 years ago

ok, i need to simplyfy this and show all working: 3^3 / sqrt3^5

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  1. BecomeMyFan=D
    • 3 years ago
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    \[3^{3}\div \sqrt{3^{5}}\]

  2. nowhereman
    • 3 years ago
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    You should use \[\sqrt{x} = x^{\frac{1}{2}}\]

  3. BecomeMyFan=D
    • 3 years ago
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    how

  4. nowhereman
    • 3 years ago
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    use power rules, and \[\frac{x}{y} = x\cdot y^{-1}\]

  5. nowhereman
    • 3 years ago
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    You do have the same bases there, so all you need no know then is what \[x^a \cdot x^b\] is

  6. BecomeMyFan=D
    • 3 years ago
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    oh, so it will be like \[3^{3-(1/5)}\] right?

  7. nowhereman
    • 3 years ago
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    nearly, but: \[\left( x^a \right) ^b = x^{a\cdot b} \]

  8. dindatc
    • 3 years ago
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    \[\sqrt{3^{5}} = 3^{1/5} . so you get 3^{3}/ \]

  9. dindatc
    • 3 years ago
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    you get 3^3 / 3^1/5

  10. Duc
    • 3 years ago
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    sqrt3

  11. dindatc
    • 3 years ago
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    hence that will be \[3^{3-1/5}\]

  12. nowhereman
    • 3 years ago
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    that is exactly the mistake! \[ \sqrt{3^5} = \left( 3^5 \right)^{\frac{1}{2}}\]

  13. dindatc
    • 3 years ago
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    oh no i think i get it wrong. that should be 3^5/2

  14. dindatc
    • 3 years ago
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    hence, its \[3^{3-}\]

  15. Duc
    • 3 years ago
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    My answer is true SQRT(3)

  16. dindatc
    • 3 years ago
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    its 3^3-5/2 = 3^1/2

  17. nowhereman
    • 3 years ago
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    And just for completeness, \[x^{\frac{1}{5}} = \sqrt[5]{x} \]

  18. dindatc
    • 3 years ago
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    the answer is \[\sqrt{3}\] right?

  19. BecomeMyFan=D
    • 3 years ago
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    duc, how do you get sqrt3?

  20. BecomeMyFan=D
    • 3 years ago
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    I know the answere, but I dont know how to get to the answer

  21. dindatc
    • 3 years ago
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    \[3^{3-5/2}\]

  22. BecomeMyFan=D
    • 3 years ago
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    ok, guys, i appreciate you helping me, i just became your fan

  23. Duc
    • 3 years ago
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    First , You 3^5=3*3^4, so SQRT(3^5)=SQRT(3)*3^2,===>(3^3)/(SQRT(3)*3^2)=3/SQRT(3)=SQRT(3) It may be TRUE

  24. BecomeMyFan=D
    • 3 years ago
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    ok, thankyou

  25. Duc
    • 3 years ago
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    No problem() good luck

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