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sparky16

  • 2 years ago

(y+1)^(2/3)=9

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  1. sparky16
    • 2 years ago
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    Solve

  2. JamesJ
    • 2 years ago
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    If (y+1)^(2/3)=9 then (y+1) = 9^(3/2) Now what is 9^(3/2) equal to?

  3. sparky16
    • 2 years ago
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    \[\sqrt[2]{9^{3}}\]

  4. JamesJ
    • 2 years ago
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    Yes, what is the value of that expression. Hint: \( 9^{3/2} = (9^{1/2})^3 = ... \)

  5. sparky16
    • 2 years ago
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    I'm not sure... \[729^{1/2}\]

  6. sparky16
    • 2 years ago
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    ???

  7. JamesJ
    • 2 years ago
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    Yes, but it's easier to write it the way I have because the square root of 9 is 3. Hence \[ 9^{3/2} = (9^{1/2})^3 = 3^3 = 27 \]

  8. sparky16
    • 2 years ago
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    where do you get the 1/2?

  9. JamesJ
    • 2 years ago
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    \[ \frac{3}{2} = 3 \times \frac{1}{2} \] hence \[ 9^{3/2} = (9^{1/2})^3 \]

  10. sparky16
    • 2 years ago
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    where do we go from there?

  11. JamesJ
    • 2 years ago
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    \[ 9^{3/2} = 9^{3 \times 1/2} = 9^{ 1/2 \times 3} = (9^{1/2})^3 = 3^3 = 27\] Now \( (y+1)^{2/3} = 9 \) is equivalent to \[ [(y+1)^{2/3}]^{3/2} = 9^{3/2} \] i.e., \[ (y+1)^1 = 27 \] i.e., \[ y + 1 = 27 \] Thus \( y = ... \)

  12. sparky16
    • 2 years ago
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    26! You are a great help! Can u help me with another problem?

  13. JamesJ
    • 2 years ago
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    Do you understand this one?

  14. sparky16
    • 2 years ago
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    yes

  15. JamesJ
    • 2 years ago
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    Post your new problem as a new problem

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