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Destinyyyy

  • one year ago

Can someone explain this to me?

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  1. Destinyyyy
    • one year ago
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    Simplify the following expression. Express the answer w/positive exponents (3xy^-1/y^3) ^-4

  2. Destinyyyy
    • one year ago
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    = (3x/ y^-1 y^3) ^-4 = (3x/y^2) ^-4 (3x/y^2) ^-4 = (3x)^-4/(y^2)^-4 = 3^-4 x^-4/y^-8 = y^8/81x^4

  3. Destinyyyy
    • one year ago
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    I know that my answer is wrong.. Where did I mess up?

  4. phi
    • one year ago
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    it's not clear what the starting expression is. \[ \left(\frac{3xy^{-1}}{y^3}\right)^{-4} \]?

  5. Destinyyyy
    • one year ago
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    Yes thats correct.. Sorry I dont get how to use the equation thing.

  6. phi
    • one year ago
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    ok so the y's inside the parens can be handled a few ways. one way: write y^-1 as 1/y \[ \left(\frac{3x}{y \cdot y^3}\right)^{-4} \\ \]

  7. phi
    • one year ago
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    the other way: y^-1 / y^3 . keep the base y. new exponent is top minus bottom exponents: -1 - 3 = -4. so y^-4 \[ \left(3xy^{-4}\right)^{-4} \\ =\left(\frac{3x}{y^4}\right)^{-4} \\\]

  8. phi
    • one year ago
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    I would "flip" the fraction and change the sign of the -4 exponent to +4 \[ \left(\frac{y^4}{3x}\right)^{4} \]

  9. phi
    • one year ago
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    can you finish ?

  10. Destinyyyy
    • one year ago
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    Um I think so... Give me a second

  11. Destinyyyy
    • one year ago
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    I have y^-16/ 3^4 x^-4 = y^-16/?

  12. Destinyyyy
    • one year ago
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    I know the final answer is y^16/(3x)^4 but im stuck on the last part

  13. phi
    • one year ago
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    if you start with \[ \left(\frac{y^4}{3x}\right)^{4} \] all the exponents stay positive

  14. phi
    • one year ago
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    if we do it "brute force" remember that \[ \left(\frac{y^4}{3x}\right)^{4} = \left(\frac{y^4}{3x}\right) \left(\frac{y^4}{3x}\right) \left(\frac{y^4}{3x}\right) \left(\frac{y^4}{3x}\right)\]

  15. phi
    • one year ago
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    and when you multiply fractions, you multiply top times top y^4 * y^4 * y^4 *y^4 = y^16 (the short way is to use the rule \[ (a^b)^c= a^{b\cdot c} \]

  16. phi
    • one year ago
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    and the bottom is \[ \left( 3x\right)^4 \]

  17. phi
    • one year ago
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    or 3^4 * x^4 or 81x^4 there are a few ways to write it.

  18. Destinyyyy
    • one year ago
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    Alright thank you!

  19. phi
    • one year ago
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    btw, when you started *** Simplify the following expression. Express the answer w/positive exponents (3xy^-1/y^3) ^-4 **** on the next post you say = (3x/ y^-1 y^3) ^-4 = (3x/y^2) ^-4 notice you changed 3xy^-1 to 3x/y^-1 when you do that (move y^-1 from the "top" to the "bottom"), you should change the sign of the exponent. you should have written: = (3x/ y^1 y^3) ^-4 and then you get = (3x/y^4) ^-4 and now you will get the correct answer

  20. Destinyyyy
    • one year ago
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    I see

  21. phi
    • one year ago
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    and to complete the thought... the other way to do this is \[ \left(\frac{y^4}{3x}\right)^{4}= \frac{\left( y^4\right)^4}{\left( 3x\right)^4}\] and then simplify the top using the rule \[ (a^b)^c= a^{bc} \] to get \(y^{4\cdot 4} = y^{16}\)

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